Exponential and Logarithmic Functions - Pre-Calculus

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Question

The amount of phosphorus present in a sample at a given time is given by the following equation:

$Q(t)=Q_0e^{-0.03t}$

Where is in days, is the amount of phosphorus after days, and is the initial amount of phosphorus at the beginning of the first day. What percent of the initial amount of phosphorus is left after days of decay?

Answer

The problem asks us for the percent that the amount of phosphorus after t days is of the original amount of phosphorus. If we think of this percentage with respect to the variables present in the equation, we can see that the following fraction expresses the amount of phosphorus after t days as a percentage of the initial amount of phosphorus:

$\frac{Q(t)}{Q_0}$

So if this is the fraction we want to solve for, we should divide both sides of the equation by to obtain this fraction on the left side of the equation:

$Q(t)=Q_0e^{-0.03t}$

$\frac{Q(t)}{Q_0}=\frac{Q_0e^{-0.03t}}{Q_0}$

$\frac{Q(t)}{Q_0}=e^{-0.03t}$

We now have the fraction we want to solve for in terms of just one variable, , for which we plug in 25 days to find the percentage of phosphorus left of the initial amount after 25 days:

$\frac{Q(25)}{Q_0}=e^{-0.03(25)}=0.47$

So after 25 days of decay, the amount of phosphorus is 47% of the initial amount.

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Question

The exponential decay of an element is given by the following function:

$Q(t)=Q_0e^{-0.04223t}$

Where is the amount of the element left after days, and is the initial amount of the element. If there are kg of the element left after days, what was the initial amount of the element?

Answer

The problem asks us for the initial amount of the element, so first let's solve our equation for :

$Q(t)=Q_0e^{-0.04223t}$

$Q_0=Q(t)e^{0.04223t}$

The problem tells us that 25 days has passed, which gives us , and it also tells us the amount left after 25 days, which gives us . Now that we have our equation for , we can plug in the given values to find the initial amount of the element:

$Q_0=Q(25)e^{0.04223(25)}=(37)e^{0.04223(25)}=106.3$ kg

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Question

The exponential decay of an element is given by the function:

$Q(t) = Q_oe^{-0.42t}$

In this function, is the amount of the element left after days, and is the initial amount of the element. If of the element is left after seven days, how much of the element was there to begin with, rounded to the nearest kilogram?

Answer

To find the initial amount, you must rearrange the equation to solve for :

$Q(t) = Q_oe^{-0.42t}$

Divide both sides by $e^{-0.42t}$ :

$Q_o = \frac{Q(t)}{e^{-0.42t}}$

Substituting in the values from the problem gives

$Q_o = \frac{25}{e^{-0.42(7)}} \approx 473:kg$

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Question

The exponential decay of an element is given by the function

$Q(t) = Q_o e^{-0.9t}$

In this function, is the amount left after days, and is the initial amount of the element. What percent of the element is left after ten days, rounded to the nearest whole percent?

Answer

To find the final percentage of the element left, we must rearrange the equation to solve for $e^{-0.9t}$ :

$\frac{Q(t)}{Q_o} = e^{-0.9t}$

Now, using the ten days as , we can solve for the percent of the element left after ten days:

$\frac{Q(10)}{Q_o} = e^{-0.9*10} \approx 0.0001 = 0.01% \approx 0%$

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Question

The exponential decay of an element is given by the function

$Q(t) = Q_o e^{-0.0327t}$

where is the amount of the element after days, and is the initial amount of the element. If of the element are left after four days, how much of the element was there initially, to the nearest tenth of a kilogram?

Answer

To solve for the initial amount, we must use rearrange the equation:

$Q_o = \frac{Q(t)}{e^{-0.0327t}}$

We now substitute the values given from the problem

$Q_o = \frac{Q(4)}{e^{-0.03274}} = \frac{3}{e^{-0.03274}} = 3.4:kg$

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Question

If you invest into a savings account which earns an interest rate per year, how much would it take for your deposit to double?

Answer

The equation of the value for this problem is

$\small R(1+i)^n=2R$ .

We can divide by R to get

$\small \small (1+i)^n=2,$ .

We want to solve for n in this case, which is the amount of years. If we use the natural log on both sides and properties of logarithms, we get

$\small \ln(1+i)^n=n\ln(1+i)=\ln2$ .

If we solve for n, we get

$\small n = \frac{\ln(2)}{\ln(1+i)}$

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Question

If you deposit into a savings account which earns a yearly interest rate, how much is in your account after two years?

Answer

Since we are investing for two years with a yearly rate of 5%, we will use the formula to calculate compound interest.

where

is the amount of money after time.

is the principal amount (initial amount).

is the interest rate.

is time.

Our amount after two years is:

$\small 100(1.05^2)=100(1.1025)=110.25$

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Question

If you deposit into a savings account which compounds interest every month, what is the expression for the amount of money in your account after years if you earn a nominal interest rate of compounded monthly?

Answer

Since is the nominal interest rate compounded monthly we write the interest term as $\frac{i}{12}$ as it is the effective monthly rate.

We compound for years which is months. Since our interest rate is compounded monthly our time needs to be in the same units thus, months will be the units of time.

Plugging this into the equation for compound interest gives us the expression:

$A=P\left(1+\frac{i}{c}\right)^{ct}$

$A=R\left(1+\frac{i}{12}\right)^{12\cdot 2}$

$A=\small \small R \left(1+\frac{i}{12}\right)^{24}$

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Question

John opens a savings account and deposits into it. This savings account gains interest per year. After years, John withdraws all the money, and deposits it into another savings account with interest per year. years later, John withdraws the money.

How much money does John have after this year period? (Assume compound interest in both accounts)

Answer

Plugging our numbers into the formula for compound interest, we have:

.

So John has about after the first three years.

After placing his money into the other savings account, he has

after more years.

So John has accumulated about .

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Question

How many years does it take for to grow into when the is deposited into a savings account that gains annual compound interest?

Answer

The correct answer is about years.

To find the number of years required, we solve the compound interest formula for .

The formula is as follows:

Substitute known values.

Divide by .

Take the natural log of both sides.

$\ln5 = \ln ((1.05)^t)$

Use the log power rule.

$\ln5 = t\ln1.03$

Divide by $\ln1.03$ .

$\frac{\ln5}{\ln1.03} = t$

use a calculator to simplify.

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Question

Suppose you took out a loan years ago that gains interest. Suppose that you haven't made any payments on it yet, and right now you owe on the loan. How much was the loan worth when you took it out?

Answer

The formula for the compund interest is as follows:

By substuting known values into the compound interest formula, we have:

$100000 = P(1.069)^{10}$ .

From here, substitute known values.

$\frac{100000}{1.069^{10}} = P$

Divide by $1.069^{10}$

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Question

There were 240mg of caffeine in the discontinued energy drink. The decay rate for caffeine in the human bloodstream is around 0.14. If Jackie drinks this energy drink around 8PM, how much caffeine will still be in her system at midnight?

Answer

Because this is a process taking place in the human body, we should use the exponential decay formula involving e:

$A=P*e^{rt}$

where A is the current amount, P is the initial amount, r is the rate of growth/decay, and t is time.

In this case, since the amount of caffeine is decreasing rather than increasing, use . Between 8PM and midnight, 4 hours pass, so use . The initial amount of caffeine is given as 240 mg, so use .

Now evaluate:

$\ A = 240\cdot e^{-0.14\cdot 4} \ \= 240\cdot e^{-0.56} \approx 240 \cdot 0.57 \ \= 137.1$

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Question

Stuff animals were a strange craze of the 90's. A Cat stuff animal with white paws sold for $6 in 1997. In 2015, the Cat will sell for $1015. What has been the approximate rate of growth for these stuff animal felines?

Answer

Use the formula for exponential growth where y is the current value, A is the initial value, r is the rate of growth, and t is time. Between 1997 and 2015, 18 years passed, so use . The stuffed animal was originally worth $6, so . It is now worth $1,015, so .

Our equation is now:

$1015 = 6*(1+r)^{18}$ divide by 6:

$169.1\overline{6}=(1+r)^{18}$ take both sides to the power of $\frac{1}{18}=0.0\overline{5}$ :

$1.329 \approx 1+r$ subtract 1

As a percent, r is about 33%.

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Question

Simplify

$\frac{9^{\frac{2}{3}}\cdot 3^{\frac{5}{3}}}{6^{2}}$

Answer

$\frac{9^{\frac{2}{3}}\cdot 3^{\frac{5}{3}}}{6^{2}}=\frac{(3^{2})^{\frac{2}{3}}\cdot3^{\frac{5}{3}}}{6^{2}}=\frac{3^{\frac{4}{3}}\cdot 3^{\frac{5}{3}}}{6^{2}}$

$=\frac{3^\frac{9}{3}}{6^2}=\frac{3^3}{6^2}=\frac{27}{36}=\frac{3}{4}$ .

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Question

Simplify the expression:

$\frac{2^\frac{5}{3}\cdot 4^\frac{2}{3}\cdot 8^\frac{1}{3}}{5^\frac{1}{3}\cdot 25^\frac{1}{3}}$ .

Answer

First, you can begin to simplfy the numerator by converting all 3 expressions into base 2.

$2^\frac{5}{3}\cdot (2^2)^\frac{4}{3}\cdot (2^3)^\frac{1}{3}$ , which simplifies to

$2^3\cdot 2 =16$

For the denominator, the same method applies. Convert the 25 into base 5, and when simplified becomes simply 5.

$5^{\frac{1}{3}}\cdot 5^{2\cdot \frac{1}{3}}$

$5^{\frac{1}{3}+\frac{2}{3}}=5^1=5$

The final simplified answer becomes:

$\frac{\text{Numerator}}{\text{Denominator}} = \frac{16}{5}$

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Question

Evaluate the following expression using knowledge of the properties of exponents:

$8^\frac{2}{3}+(2^2)^2-(3)^3\left(\frac{1}{3}\right)^3+\frac{5^{19}}{5^{17}}-647^0$

Answer

Let's work through this equation involving exponents one term at a time. The first term we see is $8^\frac{2}{3}$ , for which we can apply the following property:

$a^\frac{c}{b}=\sqrt[b]{a^c}=(\sqrt[b]{a})^c$

So if we plug our values into the formula for the property, we get:

$8^\frac{2}{3}=\sqrt[3]{8^2}=(\sqrt[3]{8})^2=(2)^2=4$

Because . Our next term is , for which we'll need the property:

$(a^b)^c=a^{bc}$

Using the values for our term, we have:

$(2^2)^2=2^4=2\cdot 2\cdot 2\cdot 2=16$

The third term of the equation is $(3)^3\left(\frac{1}{3}\right)^3$ , for which the quickest way to evaluate would be using the following property:

Using the values from our term, this gives us:

$(3)^3(\frac{1}{3})^3=(3\cdot \frac{1}{3})^3=(1)^3=1$

The next property we will need to consider for our fourth term is given below:

$\frac{a^b}{a^c}=a^{b-c}$

If we plug in the corresponding values from our term, we get:

$\frac{5^{19}}{5^{17}}=5^{19-17}=5^2=25$

Finally, our last term requires knowledge of the following simple property: Any number raised to the power of zero is 1. With this in mind, our last term becomes:

Rewriting the equation with all of the values we've just evaluated, we obtain our final answer:

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Question

Simplify the expression.

$\frac{x^{3/2}y^{7/3}}{x^3 y^{2/9}}$

Answer

Using the properties of exponents, we can either choose to subtract the exponents of the corresponding bases or rewrite the expression using negative exponents as such:

$x^{3/2}x^{-3}y^{7/3}y^{-2/9}$

Here, we combine the terms with corresponding bases by adding the exponents together to get

$x^{-3/2}y^{19/9}$

Placing the x term (since it has a negative exponent) in the denominator will result in the correct answer. It can be shown that simply subtracting the exponents of corresponding bases will result in the same answer.

$\frac{y^{19/9}}{x^{3/2}}$

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Question

Solve:

$3x^{\frac{3}{4}} = x^{\frac{1}{2}}$

Answer

To remove the fractional exponents, raise both sides to the second power and simplify:

$(3x^{\frac{3}{4}})^{2} = (x^{\frac{1}{2}})^{2}$

$9x^{\frac{3}{2}} = x$

$\frac{9x^{\frac{3}{2}}}{x} = \frac{x}{x}$

$9x^{\frac{1}{2}} = 1$

Now solve for :

$\frac{9x^{\frac{3}{2}}}{9} = \frac{1}{9}$

$x^{\frac{1}{2}} = \frac{1}{9}$

$(x^{\frac{1}{2}})^{2} = (\frac{1}{9})^{2}$

$x = \frac{1}{81}$

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Question

Solve:

$x^{\frac{1}{3}} = 2$

Answer

To remove the rational exponent, cube both sides of the equation:

$(x^{\frac{1}{3}})^{3} = (2)^{3}$

Now simplify both sides of the equation:

$x^{\frac{1}{3}\cdot 3} = 2^{3}$

$x^{1} = 2^{3}$

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Question

Evaluate the following expression and solve for .

$343=7^{15t-27}$

Answer

To solve this problem, recall that you can set exponents equal to eachother if they have the same base.

See below:

So, we have

$7^3=7^{15t-27}$

Because both sides of this equation have a base of seven, we can set the exponents equal to eachother and solve for t.

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