Exponential and Logarithmic Functions - College Algebra

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Question

On the day of a child's birth, a sum of money is to be invested into a certificate of deposit (CD) that draws annual interest compounded continuously. The plan is for the value of the CD to be at least on the child's $18^{th}$ birthday.

If the amount of money invested is to be a multiple of , what is the minimum that should be invested initially, assuming that there are no further deposits or withdrawals?

Answer

If we let be the initial amount invested and be the annual interest rate of the CD expressed as a decimal, then at the end of years, the amount of money that the CD will be worth can be determined by the formula

$A = Pe^{rt}$

Substitute , , , and solve for .

$20,000 = Pe^{0.062 \cdot 18}$

$P = \frac{20,000 }{e^{0.062 \cdot 18}} \approx \frac{20,000 }{e^{1.116}} \approx \frac{20,000 }{3.0526} \approx 6,551$

The minimum principal to be invested initially is $6,551. However, since we are looking for the multiple of $1,000 that guarantees a minimum final balance of $20,000, we round up to the nearest such multiple, which is $7,000 - the correct response.

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Question

Twelve years ago, your grandma put money into a savings account for you that earns interest annually and is continuously compounded. How much money is currently in your account if she initially deposited and you have not taken any money out?

Answer

1. Use $x=Pe^{rt}$ where is the current amount, is the interest rate, is the amount of time in years since the initial deposit, and is the amount initially deposited.

2. Solve for

$x=Pe^{rt}$

$x=10,000e^{(0.075\cdot12)}$

You currently have $24,596 in your account.

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Question

Jeffrey has won a lottery and has elected to take a $10,000 per month payment.

At the beginning of the year, Jeffrey deposits the first payment of $10,000 in an account that pays 7.6% interest annually, compounded continuously. At the very beginning of each month, he deposits another $10,000. How much will he have at the very end of the year?

Answer

The continuous compound interest formula is

$A= Pe^{rt}$ ,

where is the amount of money in the account at the end of the period, is the principal at the beginning, is the annual interest rate in decimal form, and is the number of years over which the interest accumulates. Since Jeff deposits $1,000 per month, we apply this formula twelve times, with equal to the principal at the beginning of each successive month, , and $t = \frac{1}{12} \approx 0.08333$ .

We can go ahead and calculate $e^{rt}$ , since and do not change:

$e^{rt}= e^{0.076 \cdot \frac{1}{12}} \approx e^{0.00633 } \approx 1.00635$

The formula can be rewritten as

At the beginning of January, Jeffrey deposits $10,000. At the end of January, there is

in the account.

At the beginning of February, he again deposits $10,000, so there is now

in the account.

At the end of February, there is

in the account.

Repeat addition of $10,000, then multiplication by 1.006353, ten more times to get the amount of money in the account at the end of December. This will be $125,072.98.

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Question

Sheila has won a lottery and has elected to take a $10,000 per month payment.

At the beginning of the year, Sheila deposits the first payment of $10,000 in an account that pays 7.6% interest annually, compounded monthly. At the very beginning of each month, she deposits another $10,000. How much will she have at the very end of the year?

Answer

The periodic compound interest formula is

$A= P \left (1+ \frac{r}{n} \right ) ^ {nt}$

where is the amount of money in the account at the end of the period, is the principal at the beginning, is the annual interest rate in decimal form, is the number of periods per year at which the interest is compounded, and is the number of years over which the interest accumulates.

Since Sheila deposits $1,000 per month, we apply this formula twelve times, with equal to the principal at the beginning of each successive month, , (monthly interest), and $t = \frac{1}{12} \approx 0.08333$ .

We can go ahead and calculate $\left (1+ \frac{r}{n} \right ) ^ {nt}$ , since , , and remain constant:

$\left (1+ \frac{r}{n} \right ) ^ {t} = \left (1+ \frac{0.076}{12} \right ) ^ {12 \cdot \frac{1}{12}} \approx 1.006333$ .

The formula can be rewritten as

At the beginning of January, Sheila deposits $10,000. At the end of January, there is

in the account.

At the beginning of February, she again deposits $10,000, so there is now

in the account.

At the end of February, there is

in the account.

Repeat addition of $10,000, then multiplication by 1.006333, ten more times to get the amount of money in the account at the end of December. This will be $125.056.56

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Question

An investor places $5,000 into an account that has a 5% interest rate. If the he keeps his money in the account for 57 months, and it is compounded quarterly, how much interest will he earn?

Round your answer to the nearest dollar, if needed

Answer

Use the formula for compound interest to solve: $A=P(1+\frac{r}{m})^{tm}$

"A" is the amount of interest, "P" is the initial amount invested, "r" is the interest rate, "m" is the number of times per year the interest is compounded, and "t" is the number of years the money is invested.

Plug in the appropriate values for the equation: $A=5,000(1+\frac{0.05}{4})^{(4.75)(4)}$

Because "t" is measure in years, 57 months needs to be converted to years (i.e. 57 months=4.75 years) to solve this equation

Simplify the equation: $A=5,000(1.0125)^{19}=(5,000)(1.266)=6,330$

Solution:

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Question

Which equation is equivalent to:

$\log _{4}\left( \frac{1}{2}\right)=x$

Answer

$log{_{a}}^{b}=x$ ,

$\Rightarrow a^{x}=b$

So, ${{log_{4}}^{\frac{1}{2}}}=x\Rightarrow 4^{x}=\frac{1}{2}$

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Question

$Log_{b}x=y$

What is the inverse of the log function?

Answer

This is a general formula that you should memorize. The inverse of $Log_{b}x=y$ is $b^{y}=x$ . You can use this formula to change an equation from a log function to an exponential function.

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Question

Solve:

Answer

To solve , it is necessary to know the property of .

Since and the terms cancel due to inverse operations, the answer is what's left of the term.

The answer is:

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Question

Rewrite the following expression as an exponential expression:

Answer

Rewrite the following expression as an exponential expression:

Recall the following property of logs and exponents:

Can be rewritten in the following form:

So, taking the log we are given;

We can rewrite it in the form:

$3^{14,000}=b$

So b must be a really huge number!

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Question

Convert the following logarithmic equation to an exponential equation:

$log_{13}b=147$

Answer

Convert the following logarithmic equation to an exponential equation:

$log_{13}b=147$

Recall the following:

This

Can be rewritten as

So, our given logarithm

$log_{13}b=147$

Can be rewritten as

$13^{147}=b$

Fortunately we don't need to expand, because this woud be a very large number!

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Question

Convert the following logarithmic equation to an exponential equation.

Answer

Convert the following logarithmic equation to an exponential equation.

To convert from logarithms to exponents, recall the following property:

Can be rewritten as:

So, starting with

,

We can get

$3^{156}=14x$

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Question

Solve the following:

Answer

To solve the following, you must "undo" the 5 with taking log based 5 of both sides. Thus,

$\log_5{5^x}=\log_5{125}$

$x=\log_5{125}$

The right hand side can be simplified further, as 125 is a power of 5. Thus,

$x=\log_5{125}$

$x=\log_5{5^3}$

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Question

Solve for :

$2^{x}+ 2 ^{x+2} = 171$

(Nearest hundredth)

Answer

Apply the Product of Powers Property to rewrite the second expression:

$2^{x}+ 2 ^{x+2} = 171$

$2^{x}+ 2 ^{2} \cdot 2 ^{x} = 171$

$1 \cdot 2^{x}+ 4 \cdot 2 ^{x} = 171$

Distribute out:

$(1 + 4) \cdot 2 ^{x} = 171$

$5 \cdot 2 ^{x} = 171$

Divide both sides by 5:

$\frac{5 \cdot 2 ^{x}}{5} = \frac{171}{5}$

$2 ^{x} = 34.2$

Take the natural logarithm of both sides (and note that you can use common logarithms as well):

$\ln 2 ^{x} = \ln 34.2$

Apply a property of logarithms:

$x \ln 2 = \ln 34.2$

Divide by $\ln 2$ and evaluate:

$\frac{x \ln 2}{\ln2} = \frac{\ln 34.2}{\ln2}$

$x = \frac{3.5322 }{0.6931 }$

$x \approx 5.10$

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Question

Solve for :

$3^{2x+ 1} = 9^{2x- 4 }$

(Nearest hundredth, if applicable).

Answer

$9 = 3^{2}$ , so rewrite the expression at right as a power of 3 using the Power of a Power Property:

$3^{2x+ 1} = 9^{2x- 4 }$

$3^{2x+ 1} = (3^{2} )^{2x- 4 }$

$3^{2x+ 1} = 3^{2(2x- 4 )}$

Set the exponents equal to each other and solve the resulting linear equation:

Distribute:

Subtract and 1 from both sides; we can do this simultaneously:

Divide by :

$\frac{- 2x}{-2} = \frac{-9}{-2}$

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Question

Solve the following for x:

$\log(2x+4)=1$

Answer

To solve, you must first "undo" the log. Since no base is specified, you assume it is 10. Thus, we need to take 10 to both sides.

$10^{\log(2x+4)}=10^1$

Now, simply solve for x.

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Question

Simplify the following:

$\ln{2x}+\ln{y^2}-\ln{z^7}$

Answer

To solve, you must combine the logs into 1 log, instead of three separate ones. To do this, you must remember that when adding logs, you multiply their insides, and when you subtract them, you add their insides. Therefore,

$\ln{2x}+\ln{y^2}-\ln{z^7}$

$\ln{2xy^2}-\ln{z^7}$

$\ln{\frac{2xy^2}{z^7}}$

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Question

Solve for y in the following expression:

$\log(y) + \log(5x) = 2$

Answer

To solve for y we first need to get rid of the logs.

$\log(y)+\log(5x) = 2 \rightarrow 10^{\log(y)} \cdot 10^{\log(5x)} = 100$

Then we get $y \cdot 5x = 100$ .

After that, we simply have to divide by 5x on both sides:

$y = \frac{100}{5x} = \frac{20}{x}$

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Question

Solve for .

Answer

To solve this natural logarithm equation, we must eliminate the operation. To do that, we must remember that is simply with base . So, we raise both side of the equation to the power.

$e^{ln(x^2 -3)}=e^0$

This simplifies to

. Remember that anything raised to the 0 power is 1.

Continuing to solve for x,

$x=\pm2$

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Question

Solve for .

$log_{7}(7x+7)=1$

Answer

To eliminate the operation, simply raise both side of the equation to the power because the base of the operation is 7.

$7^{log_{7}(7x+7)}=7^{1}$

This simplifies to

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Question

; $a \in (0, 1) \cup (1, \infty)$

True or false:

$\log_{a}N = \frac{\log_{b}N}{\log_{b}a}$

if and only if either or .

Answer

$\log_{a}N = \frac{\log_{b}N}{\log_{b}a}$

is a direct statement of the Change of Base Property of Logarithms. If and $a \in (0, 1) \cup (1, \infty)$ , this property holds true for any $b \in (0, 1) \cup (1, \infty)$ - not just .

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