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SAT II Math II : Exponents and Logarithms

Study concepts, example questions & explanations for SAT II Math II

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2 Diagnostic Tests 130 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #21 : Mathematical Relationships

Solve $\displaystyle ln x - ln 4 = 5$

Possible Answers:

4+e^5 $\displaystyle 4+e^5$

1024e^5 $\displaystyle 1024e^5$

$\displaystyle 1024+e^5$

4e^5 $\displaystyle 4e^5$

1024e $\displaystyle 1024e$

Correct answer:

4e^5 $\displaystyle 4e^5$

Explanation:

First, subtract the natural log terms:

$ln(\frac{x}{4})=5$ $\displaystyle ln(\frac{x}{4})=5$

Now rewrite the equation in exponential form:

$e^5=\frac{x}{4}$ $\displaystyle e^5=\frac{x}{4}$

Finally, isolate the variable:

4e^5=x $\displaystyle 4e^5=x$

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Example Question #21 : Exponents And Logarithms

Solve $log_{18} (2 - x) = log_{18} (-2x - 2)$ $\displaystyle log_{18} (2 - x) = log_{18} (-2x - 2)$

Possible Answers:

$\displaystyle 3$

$\displaystyle -2$

$\displaystyle 2$

$\displaystyle -4$

$\displaystyle 0$

Correct answer:

$\displaystyle -4$

Explanation:

We can start by canceling the logs, because they both have the same base:

$\displaystyle 2-x=-2x-2$

Now we can collect constants on one side of the equation, and variables on the other:

x=-4 $\displaystyle x=-4$

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Example Question #31 : Mathematical Relationships

Solve $\displaystyle -log_7(x - 7) + 1 = -1$ .

Possible Answers:

$\displaystyle 56$

$\displaystyle 28$

$\displaystyle 21$

$\displaystyle 49$

$\displaystyle -7$

Correct answer:

$\displaystyle 56$

Explanation:

We can start by gathering all the constants to one side of the equation:

$\displaystyle -log_7(x - 7) = -2$

Next, we can multiply by $\displaystyle -1$ to change the signs:

$\displaystyle log_7(x - 7) = 2$

Now we can rewrite the equation in exponential form:

7^2=x-7 $\displaystyle 7^2=x-7$

And finally, we can solve algebraically:

49=x-7 $\displaystyle 49=x-7$

56=x $\displaystyle 56=x$

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Example Question #22 : Exponents And Logarithms

Solve for $\displaystyle x$ :

$81^{x-3}=27^{2x}$ $\displaystyle 81^{x-3}=27^{2x}$

Possible Answers:

x=-5 $\displaystyle x=-5$

x=-4 $\displaystyle x=-4$

x=2 $\displaystyle x=2$

x=-6 $\displaystyle x=-6$

Correct answer:

x=-6 $\displaystyle x=-6$

Explanation:

In order to solve this problem, rewrite both sides of the equation in terms of raising $\displaystyle 3$ to an exponent.

Since, 81=3^4 $\displaystyle 81=3^4$ , we can write the following:

$81^{x-3}=(3^4)^{x-3}=3^{4x-12}$ $\displaystyle 81^{x-3}=(3^4)^{x-3}=3^{4x-12}$

Since 27=3^3 $\displaystyle 27=3^3$ , we can write the following:

$27^{2x}=(3^3)^{2x}=3^{6x}$ $\displaystyle 27^{2x}=(3^3)^{2x}=3^{6x}$

Now, we can solve for $\displaystyle x$ with the following equation:

$\displaystyle 6x=4x-12$

2x=-12 $\displaystyle 2x=-12$

x=-6 $\displaystyle x=-6$

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Example Question #22 : Exponents And Logarithms

Solve

$-9e^{x + 1} = 47$ $\displaystyle -9e^{x + 1} = 47$

Possible Answers:

$ln(\frac{3}{4})$ $\displaystyle ln(\frac{3}{4})$

$\displaystyle 17$

$\frac{3}{e}$ $\displaystyle \frac{3}{e}$

$\frac{e}{4}$ $\displaystyle \frac{e}{4}$

No solutions

Correct answer:

No solutions

Explanation:

The first thing we need to do is find a common base. However, because one of the bases has an $\displaystyle e$ in it (an irrational number), and the other does not, it's going to be impossible to find a common base. Therefore, the question has no solution.

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Example Question #23 : Exponents And Logarithms

Solve $log_{15} (48 + x) = log_{15} (x^2 + 3x)$ $\displaystyle log_{15} (48 + x) = log_{15} (x^2 + 3x)$

Possible Answers:

No solutions

-8, -6 $\displaystyle -8, -6$

-6, 8 $\displaystyle -6, 8$

-8, 6 $\displaystyle -8, 6$

6, 8 $\displaystyle 6, 8$

Correct answer:

-8, 6 $\displaystyle -8, 6$

Explanation:

First, we can simplify by canceling the logs, because their bases are the same:

$\displaystyle 48+x=x^2+3x$

Now we collect all the terms to one side of the equation:

$\displaystyle x^2+2x-48=0$

Factoring the expression gives:

$\displaystyle (x-6)(x+8)=0$

So our answers are:

x=-8, 6 $\displaystyle x=-8, 6$

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Example Question #23 : Exponents And Logarithms

Solve $\displaystyle 25^x = 625^x$ .

Possible Answers:

$\displaystyle 2$

$\displaystyle e$

$\displaystyle 0$

x^2 $\displaystyle x^2$

No solutions

Correct answer:

$\displaystyle 0$

Explanation:

Here, we can see that changing base isn't going to help. However, if we remember that and number raised to the $\displaystyle 0$ th power equals $\displaystyle 1$ , our solution becomes very easy.

$25^{(0)}=625^{(0)}$ $\displaystyle 25^{(0)}=625^{(0)}$

1=1 $\displaystyle 1=1$

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Example Question #24 : Exponents And Logarithms

To the nearest hundredth, solve for $\displaystyle x$ : $7^{2x} = 5 ^{x}$ $\displaystyle 7^{2x} = 5 ^{x}$ .

Possible Answers:

2.28 $\displaystyle 2.28$

0.99 $\displaystyle 0.99$

None of these

0.55 $\displaystyle 0.55$

1.27 $\displaystyle 1.27$

Correct answer:

None of these

Explanation:

Take the natural logarithm of both sides:

$7^{2x} = 5 ^{x}$ $\displaystyle 7^{2x} = 5 ^{x}$

$\ln 7^{2x} = \ln 5 ^{x}$ $\displaystyle \ln 7^{2x} = \ln 5 ^{x}$

By the Logarithm of a Power Rule the above becomes

$2x \ln 7 = x \ln 5$ $\displaystyle 2x \ln 7 = x \ln 5$

Solve for $\displaystyle x$ :

$2x \ln 7 - x \ln 5 = x \ln 5 - x \ln 5$ $\displaystyle 2x \ln 7 - x \ln 5 = x \ln 5 - x \ln 5$

$(2 \ln 7 - \ln 5) x = 0$ $\displaystyle (2 \ln 7 - \ln 5) x = 0$

$\frac{(2 \ln 7 - \ln 5) x }{2 \ln 7 - \ln 5}= \frac{0 }{2 \ln 7 - \ln 5}$ $\displaystyle \frac{(2 \ln 7 - \ln 5) x }{2 \ln 7 - \ln 5}= \frac{0 }{2 \ln 7 - \ln 5}$

x = 0 $\displaystyle x = 0$ .

This is not among the choices given.

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SAT II Math II : Exponents and Logarithms

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #21 : Mathematical Relationships

Example Question #21 : Exponents And Logarithms

Example Question #31 : Mathematical Relationships

Example Question #22 : Exponents And Logarithms

Example Question #22 : Exponents And Logarithms

Example Question #23 : Exponents And Logarithms

Example Question #23 : Exponents And Logarithms

Example Question #24 : Exponents And Logarithms

All SAT II Math II Resources

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