Algebra 2 Quiz: Solving Exponential Equations With Logarithms
20 questions · exam conditions
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Solving Exponential Equations With LogarithmsQuestion 1 of 20

Solve for tt and express the solution as a logarithm. Then use a calculator to approximate: 12⋅25t=15012\cdot 2^{5t}=150.

t=log⁡2(12.5)12≈0.304t=\dfrac{\log_2(12.5)}{12}\approx 0.304
t=log⁡(12.5)5≈0.219t=\dfrac{\log(12.5)}{5}\approx 0.219
t=log⁡2(162)5≈1.469t=\dfrac{\log_2(162)}{5}\approx 1.469
t=log⁡2(12.5)5≈0.729t=\dfrac{\log_2(12.5)}{5}\approx 0.729
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Algebra 2 Quiz

Algebra 2 Quiz: Solving Exponential Equations With Logarithms

Practice Solving Exponential Equations With Logarithms in Algebra 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Solving Exponential Equations With Logarithms, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Solve for tt and express the solution as a logarithm. Then use a calculator to approximate: 12⋅25t=15012\cdot 2^{5t}=150.

  1. t=log⁡2(12.5)12≈0.304t=\dfrac{\log_2(12.5)}{12}\approx 0.304
  2. t=log⁡(12.5)5≈0.219t=\dfrac{\log(12.5)}{5}\approx 0.219
  3. t=log⁡2(162)5≈1.469t=\dfrac{\log_2(162)}{5}\approx 1.469
  4. t=log⁡2(12.5)5≈0.729t=\dfrac{\log_2(12.5)}{5}\approx 0.729 (correct answer)
Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. When solving exponential equations like 2^t = 10 where the exponent contains the variable, logarithms are the tool that unlocks the solution: taking log base 2 of both sides gives log₂(2t2^t) = log₂(10), and using the inverse property log₂(2t2^t) = t, we get t = log₂(10). This is the exact solution! To get a decimal approximation, use your calculator with change of base: log₂(10) = ln(10)/ln(2) ≈ 3.322. For 12·2^{5t}=150, divide by 12 to isolate: 2^{5t}=12.5, take log base 2: 5t = log₂(12.5), then t = log₂(12.5)/5 ≈ 0.729. Choice A correctly isolates, takes the base-2 log, divides by 5, and approximates accurately. Choice B incorrectly uses log₂(162) instead of log₂(12.5), perhaps by misdividing 150/12 as 162 or forgetting to isolate properly, but double-check your division! Calculator strategy for non-standard bases: your calculator has log (base 10) and ln (base e) buttons, but what if you need log₂(10)? Use change of base: log₂(10) = ln(10)/ln(2) or log(10)/log(2)—both give the same answer ≈ 3.322. The formula is log_b(x) = ln(x)/ln(b) for any base b. This lets you evaluate any logarithm using just the ln button! Alternatively, leave answers in exact log form if calculator evaluation isn't required. Isolation before logarithms: ALWAYS isolate the exponential expression b^(ct) before taking logarithms. If you have 5·2^t = 40, first divide by 5 to get 2^t = 8, THEN take log. Taking log₂ of both sides of 5·2^t = 40 directly leads to log₂(5⋅2t5·2^t), which is more complex (requires log properties). Simple isolation first makes the logarithm application clean: take log of both sides when you have b^(something) = number, with the exponential alone on one side!

Question 2

Solve for tt and give both exact and approximate forms: 9⋅e0.5t=1209 \cdot e^{0.5t} = 120 (use technology to evaluate the logarithm).

  1. t=2log⁡ ⁣(1209)≈2.250t = 2 \log \! \left( \dfrac{120}{9} \right) \approx 2.250
  2. t=2ln⁡ ⁣(1209)≈5.181t = 2 \ln \! \left( \dfrac{120}{9} \right) \approx 5.181 (correct answer)
  3. t=2ln⁡ ⁣(9120)≈−5.181t = 2 \ln \! \left( \dfrac{9}{120} \right) \approx -5.181
  4. t=ln⁡(120/9)2≈1.295t = \dfrac{\ln(120/9)}{2} \approx 1.295
Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. When solving exponential equations like 2t=102^t = 10 where the exponent contains the variable, logarithms are the tool that unlocks the solution: taking log base 2 of both sides gives log2(2t)=log2(10)log_2(2^t) = log_2(10), and using the inverse property log2(2t)=tlog_2(2^t) = t, we get t=log2(10)t = log_2(10). This is the exact solution! To get a decimal approximation, use your calculator with change of base: log2(10)=ln⁡(10)/ln⁡(2)≈3.322log_2(10) = \ln(10)/\ln(2) \approx 3.322. For 9⋅e0.5t=1209 \cdot e^{0.5t} = 120, note 0.5t=t/20.5t = t/2, isolate: et/2=120/9e^{t/2} = 120/9, take ln⁡\ln: t/2=ln⁡(120/9)t/2 = \ln(120/9), t=2ln⁡(120/9)≈5.181t = 2 \ln(120/9) \approx 5.181. Choice A correctly uses natural log and multiplies by 2 for the solution. Choice B divides by 2 instead, reversing the operation—pay attention to the coefficient's position! Calculator strategy for non-standard bases: your calculator has log (base 10) and ln⁡\ln (base e) buttons, but what if you need log2(10)log_2(10)? Use change of base: log2(10)=ln⁡(10)/ln⁡(2)log_2(10) = \ln(10)/\ln(2) or log(10)/log(2)log(10)/log(2)—both give the same answer ≈3.322\approx 3.322. The formula is logb(x)=ln⁡(x)/ln⁡(b)log_b(x) = \ln(x)/\ln(b) for any base b. This lets you evaluate any logarithm using just the ln⁡\ln button! Alternatively, leave answers in exact log form if calculator evaluation isn't required. Isolation before logarithms: ALWAYS isolate the exponential expression bctb^{ct} before taking logarithms. If you have 5⋅2t=405 \cdot 2^t = 40, first divide by 5 to get 2t=82^t = 8, THEN take log. Taking log2log_2 of both sides of 5⋅2t=405 \cdot 2^t = 40 directly leads to log2(5⋅2t)log_2(5 \cdot 2^t), which is more complex (requires log properties). Simple isolation first makes the logarithm application clean: take log of both sides when you have bsomething=numberb^{something} = number, with the exponential alone on one side!

Question 3

Solve for xx and give both exact and approximate forms: 9⋅2x/5=40.59\cdot 2^{x/5}=40.5. (Express the exact answer using log⁡2\log_2.)

  1. x=5log⁡(4.5)≈3.266x=5\log(4.5)\approx 3.266
  2. x=log⁡2(4.5)5≈0.434x=\dfrac{\log_2(4.5)}{5}\approx 0.434
  3. x=5log⁡2(40.5)≈26.69x=5\log_2(40.5)\approx 26.69
  4. x=5log⁡2(4.5)≈10.85x=5\log_2(4.5)\approx 10.85 (correct answer)
Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. When solving exponential equations like 9·2^(x/5) = 40.5 where the exponent contains the variable, logarithms are the tool that unlocks the solution: first isolate the exponential by dividing both sides by 9 to get 2^(x/5) = 4.5, then take log₂ of both sides to get log₂(2^(x/5)) = log₂(4.5), and using the inverse property log₂(2^(x/5)) = x/5, we get x/5 = log₂(4.5), so x = 5·log₂(4.5). To get a decimal approximation, use change of base: log₂(4.5) = ln(4.5)/ln(2) ≈ 1.504/0.693 ≈ 2.170, so x ≈ 5(2.170) ≈ 10.85. Choice A correctly isolates the exponential term, takes log base 2, and multiplies by 5 to get x = 5·log₂(4.5) ≈ 10.85. Choice B incorrectly divides by 5 instead of multiplying, while Choice C uses log₂(40.5) without first isolating the exponential, and Choice D uses common log instead of log base 2 as requested. Calculator strategy for non-standard bases: your calculator has log (base 10) and ln (base e) buttons, but what if you need log₂(4.5)? Use change of base: log₂(4.5) = ln(4.5)/ln(2) or log(4.5)/log(2)—both give the same answer ≈ 2.170. The formula is log_b(x) = ln(x)/ln(b) for any base b. This lets you evaluate any logarithm using just the ln button!

Question 4

Use logarithms to solve the exponential equation 3⋅102x=753\cdot 10^{2x}=75 for xx. Give the exact solution (as a logarithm) and an approximate value using technology.

  1. x=ln⁡(25)2≈0.804x=\dfrac{\ln(25)}{2}\approx 0.804
  2. x=log⁡(25)≈1.398x=\log(25)\approx 1.398
  3. x=log⁡(25)2≈0.699x=\dfrac{\log(25)}{2}\approx 0.699 (correct answer)
  4. x=log⁡(75)2≈0.938x=\dfrac{\log(75)}{2}\approx 0.938
Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. The solving strategy for ab^(ct) = d: (1) Isolate the exponential by dividing both sides by a: b^(ct) = d/a, (2) Take log base b of both sides: log_b(b^(ct)) = log_b(d/a), (3) Use inverse property on left: ct = log_b(d/a), (4) Solve for variable: t = log_b(d/a)/c. This systematic approach works for any exponential equation in this form! For 3·10^{2x}=75, first divide both sides by 3 to get 10^{2x}=25, then take log base 10 of both sides: 2x = log(25), and divide by 2: x = log(25)/2 ≈ 0.699 using a calculator. Choice A correctly isolates the exponential, applies the logarithm, and evaluates to get the exact and approximate solution. Choice C forgets to divide by the 2 in the exponent, leading to x = log(25) ≈ 1.398, but always remember to solve for the variable after taking the log! Calculator strategy for non-standard bases: your calculator has log (base 10) and ln (base e) buttons, but what if you need log₂(10)? Use change of base: log₂(10) = ln(10)/ln(2) or log(10)/log(2)—both give the same answer ≈ 3.322. The formula is log_b(x) = ln(x)/ln(b) for any base b. This lets you evaluate any logarithm using just the ln button! Alternatively, leave answers in exact log form if calculator evaluation isn't required. Isolation before logarithms: ALWAYS isolate the exponential expression b^(ct) before taking logarithms. If you have 5·2^t = 40, first divide by 5 to get 2^t = 8, THEN take log. Taking log₂ of both sides of 5·2^t = 40 directly leads to log₂(5⋅2t5·2^t), which is more complex (requires log properties). Simple isolation first makes the logarithm application clean: take log of both sides when you have b^(something) = number, with the exponential alone on one side!

Question 5

Use a calculator to solve the exponential equation 9⋅e2t=1409\cdot e^{2t}=140 for tt. Express your answer in exact logarithmic form and as a decimal approximation.

  1. t=ln⁡(1409)≈2.747t=\ln\left(\dfrac{140}{9}\right)\approx 2.747
  2. t=ln⁡(9140)2≈−1.374t=\dfrac{\ln\left(\dfrac{9}{140}\right)}{2}\approx -1.374
  3. t=ln⁡(1409)2≈1.374t=\dfrac{\ln\left(\dfrac{140}{9}\right)}{2}\approx 1.374 (correct answer)
  4. t=2ln⁡(1409)≈5.494t=2\ln\left(\dfrac{140}{9}\right)\approx 5.494
Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. When solving exponential equations like 9·e^(2t) = 140 where the exponent contains the variable, logarithms are the tool that unlocks the solution: first isolate the exponential by dividing both sides by 9 to get e^(2t) = 140/9, then take ln of both sides to get ln(e^(2t)) = ln(140/9), and using the inverse property ln(e^(2t)) = 2t, we get 2t = ln(140/9), so t = ln(140/9)/2. Starting with 9·e^(2t) = 140, divide by 9 to get e^(2t) = 140/9, then take ln of both sides: 2t = ln(140/9), and divide by 2 to get t = ln(140/9)/2. Using a calculator: ln(140/9) ≈ ln(15.556) ≈ 2.747, so t ≈ 2.747/2 ≈ 1.374. Choice A correctly isolates the exponential, takes the natural logarithm, and divides by 2 to get t = ln(140/9)/2 ≈ 1.374. Choice C incorrectly multiplies by 2 instead of dividing—remember that when you have 2t = ln(140/9), you must divide both sides by 2 to solve for t! Isolation before logarithms: ALWAYS isolate the exponential expression e^(2t) before taking logarithms. If you have 9e^(2t) = 140, first divide by 9 to get e^(2t) = 140/9, THEN take ln. This makes the logarithm application clean: take ln of both sides when you have e^(something) = number!

Question 6

Use a calculator to solve the exponential equation 4⋅2t/5=304\cdot 2^{t/5}=30. Give the exact solution as a logarithm and an approximate decimal value.

  1. t=5log⁡2(152)≈14.535t=5\log_2\left(\dfrac{15}{2}\right)\approx 14.535 (correct answer)
  2. t=log⁡2(152)5≈0.581t=\dfrac{\log_2\left(\dfrac{15}{2}\right)}{5}\approx 0.581
  3. t=5log⁡2(304)≈5.814t=5\log_2\left(\dfrac{30}{4}\right)\approx 5.814
  4. t=5log⁡(152)≈4.375t=5\log\left(\dfrac{15}{2}\right)\approx 4.375
Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. When solving exponential equations like 2^t = 10 where the exponent contains the variable, logarithms are the tool that unlocks the solution: taking log base 2 of both sides gives log₂(2t2^t) = log₂(10), and using the inverse property log₂(2t2^t) = t, we get t = log₂(10). This is the exact solution! To get a decimal approximation, use your calculator with change of base: log₂(10) = ln(10)/ln(2) ≈ 3.322. To solve 4·2^{t/5}=30, divide by 4: 2^{t/5}=30/4=15/2, take log base 2: log₂(2t/52^{t/5})=log₂(15/2), simplify to t/5=log₂(15/2), and multiply by 5: t=5 log₂(15/2). Choice A correctly isolates, takes log base 2, applies the inverse property, and approximates to 14.535 using change of base on a calculator. Choice B divides by 5 instead of multiplying, resulting in t=log₂(15/2)/5 which is one-fifth the size—remember that to undo division in the exponent, you multiply when solving for t! Calculator strategy for non-standard bases: your calculator has log (base 10) and ln (base e) buttons, but what if you need log₂(10)? Use change of base: log₂(10) = ln(10)/ln(2) or log(10)/log(2)—both give the same answer ≈ 3.322. The formula is log_b(x) = ln(x)/ln(b) for any base b. This lets you evaluate any logarithm using just the ln button! Alternatively, leave answers in exact log form if calculator evaluation isn't required. Isolation before logarithms: ALWAYS isolate the exponential expression b^(ct) before taking logarithms. If you have 5·2^t = 40, first divide by 5 to get 2^t = 8, THEN take log. Taking log₂ of both sides of 5·2^t = 40 directly leads to log₂(5⋅2t5·2^t), which is more complex (requires log properties). Simple isolation first makes the logarithm application clean: take log of both sides when you have b^(something) = number, with the exponential alone on one side!

Question 7

Use logarithms to solve 15⋅22t=50015\cdot 2^{2t}=500 for tt. Express the solution as a logarithm (exact form) and then approximate using a calculator.

  1. t=log⁡2(1003)2≈2.529t=\dfrac{\log_2\left(\dfrac{100}{3}\right)}{2}\approx 2.529 (correct answer)
  2. t=log⁡2(50015)2≈1.265t=\dfrac{\log_2\left(\dfrac{500}{15}\right)}{2}\approx 1.265
  3. t=log⁡(1003)2≈0.761t=\dfrac{\log\left(\dfrac{100}{3}\right)}{2}\approx 0.761
  4. t=2log⁡2(1003)≈10.116t=2\log_2\left(\dfrac{100}{3}\right)\approx 10.116
Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. When solving exponential equations like 15·2^(2t) = 500 where the exponent contains the variable, logarithms are the tool that unlocks the solution: first isolate the exponential by dividing both sides by 15 to get 2^(2t) = 500/15 = 100/3, then take log base 2 of both sides to get log₂(2^(2t)) = log₂(100/3), and using the inverse property log₂(2^(2t)) = 2t, we get 2t = log₂(100/3), so t = log₂(100/3)/2. Starting with 15·2^(2t) = 500, divide by 15 to get 2^(2t) = 100/3, then take log₂ of both sides: 2t = log₂(100/3), and divide by 2 to get t = log₂(100/3)/2. To approximate with a calculator, use change of base: log₂(100/3) = ln(100/3)/ln(2) ≈ 3.507/0.693 ≈ 5.058, so t ≈ 5.058/2 ≈ 2.529. Choice A correctly isolates the exponential, takes log base 2, and divides by 2 to get t = log₂(100/3)/2 ≈ 2.529. Choice B incorrectly keeps the original fraction 500/15 without simplifying—while 500/15 = 100/3, the simplified form makes the calculation clearer! Calculator strategy for non-standard bases: your calculator has log (base 10) and ln (base e) buttons, but what if you need log₂(100/3)? Use change of base: log₂(100/3) = ln(100/3)/ln(2) ≈ 5.058. The formula is log_b(x) = ln(x)/ln(b) for any base b.

Question 8

Solve 6⋅10t/3=156\cdot 10^{t/3}=15 for tt. Express the solution as a logarithm and then approximate using technology.

  1. t=3log⁡(52)≈1.194t=3\log\left(\dfrac{5}{2}\right)\approx 1.194 (correct answer)
  2. t=log⁡(52)3≈0.133t=\dfrac{\log\left(\dfrac{5}{2}\right)}{3}\approx 0.133
  3. t=3log⁡(156)≈0.398t=3\log\left(\dfrac{15}{6}\right)\approx 0.398
  4. t=log⁡(52)≈0.398t=\log\left(\dfrac{5}{2}\right)\approx 0.398
Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. The solving strategy for 6·10^(t/3) = 15: (1) Isolate the exponential by dividing both sides by 6: 10^(t/3) = 15/6 = 5/2, (2) Take log base 10 of both sides: log(10^(t/3)) = log(5/2), (3) Use inverse property on left: t/3 = log(5/2), (4) Solve for variable: t = 3log(5/2). This systematic approach works for any exponential equation in this form! Starting with 6·10^(t/3) = 15, divide by 6 to get 10^(t/3) = 5/2, then take log of both sides to get t/3 = log(5/2), and multiply by 3 to get t = 3log(5/2). To get a decimal approximation, use your calculator: log(5/2) = log(2.5) ≈ 0.398, so t ≈ 3(0.398) ≈ 1.194. Choice A correctly isolates the exponential, takes the common logarithm, and multiplies by 3 to get t = 3log(5/2) ≈ 1.194. Choice B incorrectly divides by 3 instead of multiplying—when the exponent is t/3 and you get t/3 = log(5/2), you must multiply both sides by 3 to solve for t! Calculator strategy for base 10 logarithms: since we're working with 10^(t/3), we use the common logarithm (log button on calculator), which is base 10. The exact answer t = 3log(5/2) can be evaluated as 3 × log(2.5) ≈ 3 × 0.398 ≈ 1.194.

Question 9

Use logarithms to solve 5⋅23t=605\cdot 2^{3t}=60 for tt. Express your answer as a logarithm (exact form) and then approximate using a calculator.

  1. t=log⁡2(12)≈3.585t=\log_2(12)\approx 3.585
  2. t=log⁡2(60)3≈1.969t=\dfrac{\log_2(60)}{3}\approx 1.969
  3. t=log⁡(12)3≈0.360t=\dfrac{\log(12)}{3}\approx 0.360
  4. t=log⁡2(12)3≈1.195t=\dfrac{\log_2(12)}{3}\approx 1.195 (correct answer)
Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. The solving strategy for 5·2^(3t) = 60: (1) Isolate the exponential by dividing both sides by 5: 2^(3t) = 12, (2) Take log base 2 of both sides: log₂(2^(3t)) = log₂(12), (3) Use inverse property on left: 3t = log₂(12), (4) Solve for variable: t = log₂(12)/3. This systematic approach works for any exponential equation in this form! Starting with 5·2^(3t) = 60, divide by 5 to get 2^(3t) = 12, then take log₂ of both sides to get 3t = log₂(12), and finally divide by 3 to get t = log₂(12)/3. To approximate with a calculator, use change of base: log₂(12) = ln(12)/ln(2) ≈ 3.585, so t ≈ 3.585/3 ≈ 1.195. Choice A correctly isolates the exponential, takes log base 2, and divides by 3 to get t = log₂(12)/3 ≈ 1.195. Choice B incorrectly uses 60 instead of 12 after isolation—remember to divide 60 by 5 first to get 2^(3t) = 12, not 2^(3t) = 60! Calculator strategy for non-standard bases: your calculator has log (base 10) and ln (base e) buttons, but what if you need log₂(12)? Use change of base: log₂(12) = ln(12)/ln(2) or log(12)/log(2)—both give the same answer ≈ 3.585. The formula is log_b(x) = ln(x)/ln(b) for any base b.

Question 10

Solve the exponential equation 3⋅102x=753\cdot 10^{2x}=75 for xx. Give the exact solution (as a logarithm) and an approximate value using technology.

  1. x=log⁡(25)2≈0.699x=\dfrac{\log(25)}{2}\approx 0.699 (correct answer)
  2. x=log⁡(75)2≈0.938x=\dfrac{\log(75)}{2}\approx 0.938
  3. x=log⁡(25)≈1.398x=\log(25)\approx 1.398
  4. x=ln⁡(25)2≈1.609x=\dfrac{\ln(25)}{2}\approx 1.609
Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. The solving strategy for ab^(ct) = d: (1) Isolate the exponential by dividing both sides by a: b^(ct) = d/a, (2) Take log base b of both sides: log_b(b^(ct)) = log_b(d/a), (3) Use inverse property on left: ct = log_b(d/a), (4) Solve for variable: t = log_b(d/a)/c. This systematic approach works for any exponential equation in this form! To solve 3·10^{2x}=75, first divide both sides by 3 to isolate the exponential: 10^{2x}=25, then take the common logarithm (base 10) of both sides: log(102x10^{2x})=log(25), which simplifies to 2x=log(25) using the inverse property, and finally divide by 2: x=log(25)/2. Choice A correctly isolates and takes the logarithm to get the exact solution as log(25)/2 and approximates it to 0.699 using technology. Choice C forgets to divide by 2 after taking the log, resulting in x=log(25) which is twice as large as it should be—remember to solve fully for the variable after applying the logarithm! Calculator strategy for non-standard bases: your calculator has log (base 10) and ln (base e) buttons, but what if you need log₂(10)? Use change of base: log₂(10) = ln(10)/ln(2) or log(10)/log(2)—both give the same answer ≈ 3.322. The formula is log_b(x) = ln(x)/ln(b) for any base b. This lets you evaluate any logarithm using just the ln button! Alternatively, leave answers in exact log form if calculator evaluation isn't required. Isolation before logarithms: ALWAYS isolate the exponential expression b^(ct) before taking logarithms. If you have 5·2^t = 40, first divide by 5 to get 2^t = 8, THEN take log. Taking log₂ of both sides of 5·2^t = 40 directly leads to log₂(5⋅2t5·2^t), which is more complex (requires log properties). Simple isolation first makes the logarithm application clean: take log of both sides when you have b^(something) = number, with the exponential alone on one side!

Question 11

Solve and evaluate: 2⋅e0.5t=202\cdot e^{0.5t}=20. Write tt in exact logarithmic form and then approximate using technology.

  1. t=2ln⁡(10)≈4.605t=2\ln(10)\approx 4.605 (correct answer)
  2. t=ln⁡(10)≈2.303t=\ln(10)\approx 2.303
  3. t=2ln⁡(20)≈5.991t=2\ln(20)\approx 5.991
  4. t=ln⁡(5)≈1.609t=\ln(5)\approx 1.609
Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. The solving strategy for ab^(ct) = d: (1) Isolate the exponential by dividing both sides by a: b^(ct) = d/a, (2) Take log base b of both sides: log_b(b^(ct)) = log_b(d/a), (3) Use inverse property on left: ct = log_b(d/a), (4) Solve for variable: t = log_b(d/a)/c. This systematic approach works for any exponential equation in this form! To solve 2·e^{0.5t}=20, divide by 2: e^{0.5t}=10, take natural log: ln(e0.5te^{0.5t})=ln(10), simplify to 0.5t=ln(10), and multiply by 2: t=2 ln(10). Choice A correctly isolates, takes the natural logarithm, applies the inverse property since ln(eye^y)=y, and approximates to 4.605. Choice B forgets to multiply by 2, resulting in t=ln(10) which is half the correct value—be sure to solve for t by dividing by the coefficient 0.5, which is the same as multiplying by 2! Calculator strategy for non-standard bases: your calculator has log (base 10) and ln (base e) buttons, but what if you need log₂(10)? Use change of base: log₂(10) = ln(10)/ln(2) or log(10)/log(2)—both give the same answer ≈ 3.322. The formula is log_b(x) = ln(x)/ln(b) for any base b. This lets you evaluate any logarithm using just the ln button! Alternatively, leave answers in exact log form if calculator evaluation isn't required. Isolation before logarithms: ALWAYS isolate the exponential expression b^(ct) before taking logarithms. If you have 5·2^t = 40, first divide by 5 to get 2^t = 8, THEN take log. Taking log₂ of both sides of 5·2^t = 40 directly leads to log₂(5⋅2t5·2^t), which is more complex (requires log properties). Simple isolation first makes the logarithm application clean: take log of both sides when you have b^(something) = number, with the exponential alone on one side!

Question 12

Solve for tt: 12⋅2t/4=10012\cdot 2^{t/4}=100. Give the exact solution as a logarithm and then approximate using technology.

  1. t=4log⁡2(253)≈12.235t=4\log_2\left(\dfrac{25}{3}\right)\approx 12.235 (correct answer)
  2. t=log⁡2(253)4≈0.765t=\dfrac{\log_2\left(\dfrac{25}{3}\right)}{4}\approx 0.765
  3. t=4log⁡(253)≈3.684t=4\log\left(\dfrac{25}{3}\right)\approx 3.684
  4. t=log⁡2(253)≈3.059t=\log_2\left(\dfrac{25}{3}\right)\approx 3.059
Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. The solving strategy for 12·2^(t/4) = 100: (1) Isolate the exponential by dividing both sides by 12: 2^(t/4) = 100/12 = 25/3, (2) Take log base 2 of both sides: log₂(2^(t/4)) = log₂(25/3), (3) Use inverse property on left: t/4 = log₂(25/3), (4) Solve for variable: t = 4log₂(25/3). This systematic approach works for any exponential equation in this form! Starting with 12·2^(t/4) = 100, divide by 12 to get 2^(t/4) = 25/3, then take log₂ of both sides to get t/4 = log₂(25/3), and multiply by 4 to get t = 4log₂(25/3). To approximate with a calculator, use change of base: log₂(25/3) = ln(25/3)/ln(2) ≈ 2.120/0.693 ≈ 3.059, so t ≈ 4(3.059) ≈ 12.235. Choice A correctly isolates the exponential, takes log base 2, and multiplies by 4 to get t = 4log₂(25/3) ≈ 12.235. Choice B incorrectly divides by 4 instead of multiplying—when the exponent is t/4 and you get t/4 = log₂(25/3), you must multiply both sides by 4 to solve for t! Calculator strategy for non-standard bases: your calculator has log (base 10) and ln (base e) buttons, but what if you need log₂(25/3)? Use change of base: log₂(25/3) = ln(25/3)/ln(2) ≈ 3.059. Always multiply by the coefficient when the variable appears as t/4 in the exponent!

Question 13

Solve for xx: 5⋅23x=605\cdot 2^{3x}=60. Express your solution as a logarithm (base 2) and then approximate using a calculator.

  1. x=log⁡2(12)3≈1.195x=\dfrac{\log_2(12)}{3}\approx 1.195 (correct answer)
  2. x=log⁡(12)3≈0.360x=\dfrac{\log(12)}{3}\approx 0.360
  3. x=log⁡2(12)≈3.585x=\log_2(12)\approx 3.585
  4. x=log⁡2(60)3≈1.969x=\dfrac{\log_2(60)}{3}\approx 1.969
Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. The solving strategy for 5·2^(3x) = 60: (1) Isolate the exponential by dividing both sides by 5: 2^(3x) = 12, (2) Take log base 2 of both sides: log₂(2^(3x)) = log₂(12), (3) Use inverse property on left: 3x = log₂(12), (4) Solve for variable: x = log₂(12)/3. This systematic approach works for any exponential equation in this form! Starting with 5·2^(3x) = 60, divide by 5 to get 2^(3x) = 12, then take log₂ of both sides: log₂(2^(3x)) = log₂(12), which simplifies to 3x = log₂(12), so x = log₂(12)/3. To evaluate with a calculator, use change of base: log₂(12) = ln(12)/ln(2) ≈ 3.585/1.386 ≈ 3.585, so x ≈ 3.585/3 ≈ 1.195. Choice A correctly isolates the exponential, takes log base 2, and divides by 3 to get x = log₂(12)/3 ≈ 1.195. Choice B incorrectly uses log₂(60) without first isolating the exponential, while Choice D uses common log instead of log base 2 as requested. Calculator strategy for non-standard bases: your calculator has log (base 10) and ln (base e) buttons, but what if you need log₂(12)? Use change of base: log₂(12) = ln(12)/ln(2) or log(12)/log(2)—both give the same answer ≈ 3.585. The formula is log_b(x) = ln(x)/ln(b) for any base b. This lets you evaluate any logarithm using just the ln button!

Question 14

What is tt in the equation 7⋅10t/3=507\cdot 10^{t/3}=50? Express tt as a logarithm and evaluate using technology.

  1. t=log⁡(507)3≈0.629t=\log\left(\dfrac{50}{7}\right)^3\approx 0.629
  2. t=3log⁡(750)≈−2.571t=3\log\left(\dfrac{7}{50}\right)\approx -2.571
  3. t=3log⁡(507)≈2.571t=3\log\left(\dfrac{50}{7}\right)\approx 2.571 (correct answer)
  4. t=log⁡(507)≈0.857t=\log\left(\dfrac{50}{7}\right)\approx 0.857
Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. The solving strategy for ab^(ct) = d: (1) Isolate the exponential by dividing both sides by a: b^(ct) = d/a, (2) Take log base b of both sides: log_b(b^(ct)) = log_b(d/a), (3) Use inverse property on left: ct = log_b(d/a), (4) Solve for variable: t = log_b(d/a)/c. This systematic approach works for any exponential equation in this form! To solve 7·10^{t/3}=50, divide by 7: 10^{t/3}=50/7, take common log: log(10t/310^{t/3})=log(50/7), simplify to t/3=log(50/7), and multiply by 3: t=3 log(50/7). Choice A correctly isolates the exponential, takes the common logarithm, uses the inverse property, and provides the exact form with an approximation. Choice B skips multiplying by 3, giving t=log(50/7) which is one-third the correct value—don't forget to multiply by the reciprocal of the exponent's coefficient! Calculator strategy for non-standard bases: your calculator has log (base 10) and ln (base e) buttons, but what if you need log₂(10)? Use change of base: log₂(10) = ln(10)/ln(2) or log(10)/log(2)—both give the same answer ≈ 3.322. The formula is log_b(x) = ln(x)/ln(b) for any base b. This lets you evaluate any logarithm using just the ln button! Alternatively, leave answers in exact log form if calculator evaluation isn't required. Isolation before logarithms: ALWAYS isolate the exponential expression b^(ct) before taking logarithms. If you have 5·2^t = 40, first divide by 5 to get 2^t = 8, THEN take log. Taking log₂ of both sides of 5·2^t = 40 directly leads to log₂(5⋅2t5·2^t), which is more complex (requires log properties). Simple isolation first makes the logarithm application clean: take log of both sides when you have b^(something) = number, with the exponential alone on one side!

Question 15

Use logarithms to solve the equation 2.5⋅et/4=402.5\cdot e^{t/4}=40 for tt. Give the exact solution and an approximate value using a calculator.

  1. t=ln⁡(16)≈2.773t=\ln(16)\approx 2.773
  2. t=4ln⁡(42.5)≈15.003t=4\ln(42.5)\approx 15.003
  3. t=ln⁡(16)4≈0.693t=\dfrac{\ln(16)}{4}\approx 0.693
  4. t=4ln⁡(16)≈11.090t=4\ln(16)\approx 11.090 (correct answer)
Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. The solving strategy for ab^(ct) = d: (1) Isolate the exponential by dividing both sides by a: b^(ct) = d/a, (2) Take log base b of both sides: log_b(b^(ct)) = log_b(d/a), (3) Use inverse property on left: ct = log_b(d/a), (4) Solve for variable: t = log_b(d/a)/c. This systematic approach works for any exponential equation in this form! To solve 2.5·e^{t/4}=40, divide by 2.5: e^{t/4}=16, take natural log: ln(et/4e^{t/4})=ln(16), simplify to t/4=ln(16), and multiply by 4: t=4 ln(16). Choice A correctly isolates, takes the natural logarithm, uses the inverse property, and approximates to 11.090 using technology. Choice B divides by 4 instead of multiplying, resulting in t=ln(16)/4 which is one-fourth the value—always multiply by the reciprocal of the exponent's coefficient! Calculator strategy for non-standard bases: your calculator has log (base 10) and ln (base e) buttons, but what if you need log₂(10)? Use change of base: log₂(10) = ln(10)/ln(2) or log(10)/log(2)—both give the same answer ≈ 3.322. The formula is log_b(x) = ln(x)/ln(b) for any base b. This lets you evaluate any logarithm using just the ln button! Alternatively, leave answers in exact log form if calculator evaluation isn't required. Isolation before logarithms: ALWAYS isolate the exponential expression b^(ct) before taking logarithms. If you have 5·2^t = 40, first divide by 5 to get 2^t = 8, THEN take log. Taking log₂ of both sides of 5·2^t = 40 directly leads to log₂(5⋅2t5·2^t), which is more complex (requires log properties). Simple isolation first makes the logarithm application clean: take log of both sides when you have b^(something) = number, with the exponential alone on one side!

Question 16

Use logarithms to solve 7⋅25t=1207\cdot 2^{5t}=120 for tt. Express the solution as a logarithm (base 2) and evaluate using a calculator.

  1. t=log⁡2(1207)5≈0.820t=\dfrac{\log_2\left(\frac{120}{7}\right)}{5}\approx 0.820 (correct answer)
  2. t=log⁡2(7120)5≈−0.820t=\dfrac{\log_2\left(\frac{7}{120}\right)}{5}\approx -0.820
  3. t=log⁡2(1207)7≈0.586t=\dfrac{\log_2\left(\frac{120}{7}\right)}{7}\approx 0.586
  4. t=log⁡(1207)5≈0.246t=\dfrac{\log\left(\frac{120}{7}\right)}{5}\approx 0.246
Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. The solving strategy for 7·2^(5t) = 120 follows our systematic approach: (1) Isolate the exponential by dividing both sides by 7: 2^(5t) = 120/7, (2) Take log base 2 of both sides: log₂(2^(5t)) = log₂(120/7), (3) Use inverse property on left: 5t = log₂(120/7), (4) Solve for variable: t = log₂(120/7)/5. This systematic approach works for any exponential equation in this form! Starting with 7·2^(5t) = 120, I divide by 7 to get 2^(5t) = 120/7 ≈ 17.143, then take log₂ of both sides: log₂(2^(5t)) = log₂(120/7), which simplifies to 5t = log₂(120/7) using the inverse property, and finally divide by 5: t = log₂(120/7)/5. Using change of base formula: log₂(120/7) = ln(120/7)/ln(2) ≈ 2.842/0.693 ≈ 4.100, so t ≈ 4.100/5 ≈ 0.820. Choice A correctly isolates the exponential, takes log base 2, and divides by 5 to get t = log₂(120/7)/5 ≈ 0.820. Choice D incorrectly uses common log (base 10) instead of log base 2, which gives log(120/7)/5 ≈ 1.234/5 ≈ 0.246—this is wrong because we need the same base as our exponential expression. Calculator strategy for non-standard bases: your calculator has log (base 10) and ln (base e) buttons, but what if you need log₂(120/7)? Use change of base: log₂(120/7) = ln(120/7)/ln(2) or log(120/7)/log(2)—both give the same answer ≈ 4.100. The formula is log_b(x) = ln(x)/ln(b) for any base b. This lets you evaluate any logarithm using just the ln button!

Question 17

Solve for xx and evaluate using technology: 8⋅2x/2=708\cdot 2^{x/2}=70. Express the exact solution as a logarithm.

  1. x=log⁡2(354)2≈1.565x=\dfrac{\log_2\left(\dfrac{35}{4}\right)}{2}\approx 1.565
  2. x=2log⁡2(708)≈3.129x=2\log_2\left(\dfrac{70}{8}\right)\approx 3.129
  3. x=2log⁡2(354)≈6.258x=2\log_2\left(\dfrac{35}{4}\right)\approx 6.258 (correct answer)
  4. x=2log⁡(354)≈1.792x=2\log\left(\dfrac{35}{4}\right)\approx 1.792
Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. When solving exponential equations like 2^t = 10 where the exponent contains the variable, logarithms are the tool that unlocks the solution: taking log base 2 of both sides gives log₂(2t2^t) = log₂(10), and using the inverse property log₂(2t2^t) = t, we get t = log₂(10). This is the exact solution! To get a decimal approximation, use your calculator with change of base: log₂(10) = ln(10)/ln(2) ≈ 3.322. To solve 8·2^{x/2}=70, divide by 8: 2^{x/2}=70/8=35/4, take log base 2: log₂(2x/22^{x/2})=log₂(35/4), simplify to x/2=log₂(35/4), and multiply by 2: x=2 log₂(35/4). Choice A correctly isolates the exponential, takes log base 2, applies the inverse property, and approximates to 6.258 using change of base. Choice B divides by 2 instead of multiplying, giving x=log₂(35/4)/2 which is half the correct value—remember to multiply by 2 to undo the division in the exponent! Calculator strategy for non-standard bases: your calculator has log (base 10) and ln (base e) buttons, but what if you need log₂(10)? Use change of base: log₂(10) = ln(10)/ln(2) or log(10)/log(2)—both give the same answer ≈ 3.322. The formula is log_b(x) = ln(x)/ln(b) for any base b. This lets you evaluate any logarithm using just the ln button! Alternatively, leave answers in exact log form if calculator evaluation isn't required. Isolation before logarithms: ALWAYS isolate the exponential expression b^(ct) before taking logarithms. If you have 5·2^t = 40, first divide by 5 to get 2^t = 8, THEN take log. Taking log₂ of both sides of 5·2^t = 40 directly leads to log₂(5⋅2t5·2^t), which is more complex (requires log properties). Simple isolation first makes the logarithm application clean: take log of both sides when you have b^(something) = number, with the exponential alone on one side!

Question 18

Use a calculator to solve the exponential equation 6⋅20.25t=906\cdot 2^{0.25t}=90 for tt. Give the exact solution as a logarithm and an approximate value.

  1. t=log⁡2(15)0.25≈15.628t=\dfrac{\log_{2}(15)}{0.25}\approx 15.628 (correct answer)
  2. t=0.25log⁡2(15)≈0.977t=0.25\log_{2}(15)\approx 0.977
  3. t=log⁡2(90)0.25≈26.196t=\dfrac{\log_{2}(90)}{0.25}\approx 26.196
  4. t=log⁡(15)0.25≈4.705t=\dfrac{\log(15)}{0.25}\approx 4.705
Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. To solve 6·2^(0.25t) = 90, first isolate the exponential by dividing by 6: 2^(0.25t) = 90/6 = 15. Taking log base 2 of both sides: log₂(2^(0.25t)) = log₂(15), using the inverse property: 0.25t = log₂(15), and dividing by 0.25 (which is the same as multiplying by 4): t = log₂(15)/0.25 = 4·log₂(15). Using change of base: log₂(15) = ln(15)/ln(2) ≈ 2.708/0.693 ≈ 3.907, so t ≈ 4·3.907 ≈ 15.628. Choice A correctly isolates the exponential and divides by the coefficient 0.25 to get t = log₂(15)/0.25 ≈ 15.628. Choice B incorrectly multiplies by 0.25 instead of dividing, while Choice D uses log base 10 instead of log base 2. Remember that dividing by 0.25 is the same as multiplying by 4—this is why the answer is relatively large. Always double-check: if t ≈ 15.628, then 0.25t ≈ 3.907, and 2^3.907 ≈ 15, so 6·15 = 90 ✓

Question 19

What is xx in 2⋅25x=702\cdot 2^{5x}=70? Express the solution as a logarithm (exact form) and then approximate using a calculator.

  1. x=log⁡2(70)5≈1.129x=\dfrac{\log_2(70)}{5}\approx 1.129
  2. x=5log⁡2(35)≈25.651x=5\log_2(35)\approx 25.651
  3. x=log⁡2(35)5≈1.026x=\dfrac{\log_2(35)}{5}\approx 1.026 (correct answer)
  4. x=log⁡(35)5≈0.309x=\dfrac{\log(35)}{5}\approx 0.309
Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. When solving exponential equations like 2·2^(5x) = 70 where the exponent contains the variable, logarithms are the tool that unlocks the solution: first isolate the exponential by dividing both sides by 2 to get 2^(5x) = 35, then take log base 2 of both sides to get log₂(2^(5x)) = log₂(35), and using the inverse property log₂(2^(5x)) = 5x, we get 5x = log₂(35), so x = log₂(35)/5. Starting with 2·2^(5x) = 70, divide by 2 to get 2^(5x) = 35, then take log₂ of both sides: 5x = log₂(35), and divide by 5 to get x = log₂(35)/5. To approximate with a calculator, use change of base: log₂(35) = ln(35)/ln(2) ≈ 3.555/0.693 ≈ 5.129, so x ≈ 5.129/5 ≈ 1.026. Choice A correctly isolates the exponential, takes log base 2, and divides by 5 to get x = log₂(35)/5 ≈ 1.026. Choice B incorrectly uses 70 instead of 35 after isolation—remember to divide 70 by 2 first to get 2^(5x) = 35, not 2^(5x) = 70! Calculator strategy for non-standard bases: your calculator has log (base 10) and ln (base e) buttons, but what if you need log₂(35)? Use change of base: log₂(35) = ln(35)/ln(2) or log(35)/log(2)—both give the same answer ≈ 5.129.

Question 20

Solve and evaluate using technology: 12⋅10t/4=50012\cdot 10^{t/4}=500. Give tt in exact logarithmic form and as a decimal approximation.

  1. t=4log⁡(50012)≈6.48t=4\log\left(\frac{500}{12}\right)\approx 6.48 (correct answer)
  2. t=log⁡(50012)4≈0.405t=\dfrac{\log\left(\frac{500}{12}\right)}{4}\approx 0.405
  3. t=4log⁡(12500)≈−6.48t=4\log\left(\frac{12}{500}\right)\approx -6.48
  4. t=4ln⁡(50012)≈14.92t=4\ln\left(\frac{500}{12}\right)\approx 14.92
Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. When solving exponential equations like 12·10^(t/4) = 500 where the exponent contains t/4, logarithms unlock the solution: first isolate the exponential by dividing both sides by 12 to get 10^(t/4) = 500/12 ≈ 41.667, then take log base 10 of both sides to get log(10^(t/4)) = log(500/12), and using the inverse property log(10^(t/4)) = t/4, we get t/4 = log(500/12), so t = 4·log(500/12). To solve 12·10^(t/4) = 500 systematically: divide by 12 to get 10^(t/4) = 41.667, take common log of both sides: log(10^(t/4)) = log(41.667), apply the inverse property to get t/4 = log(41.667) ≈ 1.620, and multiply by 4: t = 4·log(41.667) = 4·log(500/12) ≈ 4·1.620 ≈ 6.48. Choice A correctly isolates the exponential, takes the common logarithm, and multiplies by 4 to get t = 4·log(500/12) ≈ 6.48. Choice D uses natural log (ln) instead of common log, giving t = 4·ln(500/12) ≈ 4·3.730 ≈ 14.92, which is wrong because we need log base 10 to match our base 10 exponential—using ln would require additional conversion. Isolation before logarithms: ALWAYS isolate the exponential expression 10^(t/4) before taking logarithms. If you have 12·10^(t/4) = 500, first divide by 12 to get 10^(t/4) = 41.667, THEN take log. When the variable appears as t/4 in the exponent, remember that after taking the logarithm you get t/4 = log(41.667), so you must multiply by 4 to solve for t—don't divide by 4!