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College Algebra : Exponential and Logarithmic Functions

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Example Questions

Example Question #21 : Exponential And Logarithmic Functions

Condense this logarithm: $\frac{1}{3}[2\ln (x+3)+\ln x-\ln(x^2-1)]$

Possible Answers:

$\ln\sqrt[3]{\frac{(x+3)^2}{x^2-1}}$

$\ln\sqrt[3]{\frac{x(x+3)^2}{x^2-1}}$

None of these

$\ln\sqrt[3]{\frac{x(x-3)^2}{x^2-1}}$

$\ln\sqrt[3]{\frac{x(x-3)^2}{x^2+1}}$

Correct answer:

$\ln\sqrt[3]{\frac{x(x+3)^2}{x^2-1}}$

Explanation:

We condense this logarithm based on the following properties:

$\ln \frac{u}{v}= \ln u-\ln v$

$\ln u^{n}= n \ln u$

$\ln uv=\ln u+\ln v$

$\frac{1}{3}[2\ln (x+3)+\ln x-\ln(x^2-1)]\rightarrow \frac{1}{3}[\ln(x+3)^2+\ln x-\ln(x^2-1)]\rightarrow \frac{1}{3}[ \ln x(x+3)^2-\ln (x^2-1)]\rightarrow$

$\frac{1}{3}[\ln\frac{x(x+3)^2}{x^2-1}]\rightarrow\boldsymbol{ \ln \sqrt[3]{\frac{x(x+3)^2}{x^2-1}}}$

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Example Question #1 : Natural Log

$5^{x}=2e^5$

Solve for .

Possible Answers:

$x = \frac{\textup{ln}2+5}{\textup{ln}5}$

$x = \frac{\textup{ln}5}{\textup{ln}2+5}$

x=5

$x = \frac{\textup{ln}2}{\textup{ln}5}$

Correct answer:

$x = \frac{\textup{ln}2+5}{\textup{ln}5}$

Explanation:

The first thing we notice about this problem is that is an exponent. This should be an immediate reminder: use logs!

The question is, which base should we choose for the log? We should use the natural log (log base e) because the right-hand side of the equation already has e as a base of an exponent. As you will see, things cancel out more nicely this way.

$5^{x}=2e^5$

Take the natural log of both sides:

$\textup{ln}(5^x) = \textup{ln}(2e^5)$

Rewrite the right-hand side of the equation using the product rule for logs:

$\textup{ln}5^x = \textup{ln}2+\textup{ln}e^5$

Now rewrite the whole equation after bringing down those exponents.

$x\textup{ln}5 = \textup{ln}2+5\textup{ln}e$

$\textup{ln}e$ is the same thing as $\textup{log}_{e}e$ , which equals 1.

$x\textup{ln}5 = \textup{ln}2+5$

Now we just divide by $\textup{ln}5$ on both sides to isolate .

$x = \frac{\textup{ln}2+5}{\textup{ln}5}$

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Example Question #22 : Exponential And Logarithmic Functions

Solve $6e^{-x}+1=3$ . Round to the nearest thousandth.

Possible Answers:

x=1.211

x=-2.001

x=-1.099

x=0.089

Correct answer:

x=-1.099

Explanation:

The original equation is:

$6e^{-x}+1=3$

Subtract from both sides:

$6e^{-x}=2$

Divde both sides by :

$e^{-x}=\frac{2}{6}=\frac{1}{3}$

Take the natural logarithm of both sides:

$-x=\ln \frac{1}{3}$

Divde both sides by and use a calculator to get:

x=-1.099

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Example Question #2 : Natural Log

Solve for :

ln(x+3)=3 .

If necessary, round to the nearest tenth.

Possible Answers:

No solution

x=17.1

x=3

x=5.7

x=2.7

Correct answer:

x=17.1

Explanation:

Give both sides the same base, using e:

$e^{ln(x+3)}=e^3$ .

Because e and ln cancel each other out, x+3=e^3 .

Solve for x and round to the nearest tenth:

x=e^3-3

$x\approx17.1$

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Example Question #1 : Solving Exponential Functions

Solve the following for x:

$e^{x+2}=3$

Possible Answers:

$\ln3-2$

e^3-2

$\ln2-3$

$\ln1$

Correct answer:

$\ln3-2$

Explanation:

To solve this exponential function, you must first "undo" the e by taking the natural log of both sides. Thus,

$\ln(e^{x+2})=\ln3$

$x+2=\ln3$

Then, to isolate x, you must subtract two from both sides. Thus,

$x=\ln3-2$

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Example Question #2 : Solving Exponential Functions

Solve the following for x:

$2e^{4x}=4$

Possible Answers:

$\frac{4}{\ln2}$

$\frac{2}{\ln4}$

$\frac{\ln4}{2}$

$\frac{\ln2}{4}$

Correct answer:

$\frac{\ln2}{4}$

Explanation:

To solve, you must isolate x. The first two steps are to divide both sides by 2 and then take the natural log of both sides.

$e^{4x}=2$

$\ln{e^{4x}}=\ln2$

$4x=\ln2$

Finally, you must divide both sides by 4.

$\frac{4x}{4}=\frac{\ln2}{4}$

$x=\frac{\ln2}{4}$

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Example Question #3 : Solving Exponential Functions

Give the solution set for the exponential equation shown below for the following cases:

Case 1,

a= b

Case 2

$a \neq b$

$a^{10x-1}=b^{10x-1}$

Possible Answers:

Case 1:

There are no solutions,

$x =\left \{ \varnothing \right \}$

Case 2:

$x = \left \{ 0\right \}$

Case 1:

All real values of are solutions:

$x=\left \{x| -\infty < x <\infty \right \}$

Case 2:

All real values of are solutions:

$x=\left \{x| -\infty < x <\infty \right \}$

Case 1:

$x = \left \{ \frac{1}{10}\right \}$

Case 2:

$x = \left \{ \frac{1}{10}\right \}$

Case 1:

$x=\left \{x| -\infty < x <\infty \right \}$

Case 2:

There is not enough information to find if we do not know the values of and b.

Case 1:

All real values of are solutions:

$x=\left \{x| -\infty < x <\infty \right \}$

Case 2:

$x = \left \{ \frac{1}{10}\right \}$

Correct answer:

Case 1:

All real values of are solutions:

$x=\left \{x| -\infty < x <\infty \right \}$

Case 2:

$x = \left \{ \frac{1}{10}\right \}$

Explanation:

Case 1,

a= b

$a^{10x-1}=b^{10x-1}$

$a^{10x-1}=a^{10x-1}$

10x-1 = 10x-1

x=x

This is true for all real values of , therefore,

$x=\left \{x| -\infty < x <\infty \right \}$

Case 2

$a \neq b$

$a^{10x-1}=b^{10x-1}$

Take the natural logarithm of both sides (note that we could use any logarithm but it's convenient to just choose the natural logarithm).

$\ln\left(a^{10x-1} \right ) = \ln\left[b^{10x-1}\right]$

Use the rule for pulling out exponents $\ln(u^c)=c\ln(u)$ .

$(10x-1)\ln(a) = (10x-1)\ln(b)$

Note that if you were to divide out by 10x-1 , you would obtain $\ln(a)=\ln(b)$ which would imply a= b which is not true for this case. We are solving for $a\neq b$ .

Expand with the distributive property,

$10x\ln(a)-\ln(a)=10x\ln(b)-\ln(b)$

Collect and isolate terms with onto one side of the equation,

$10x\ln(a)-10x\ln(b)=\ln(a)-\ln(b)$

Factor out 10x (you could just factor out and leave the in front of the logarithms but it's easier to see the solution writing it this way).

$10x[\ln(a)-\ln(b)]=\ln(a)-\ln(b)$

$10x = \frac{\ln(a)-\ln(b)}{\ln(a)-\ln(b)}$

10x =1

$x = \frac{1}{10}$

What's remarkable about this solution is that it was obtained without specifying any value for or . The solution is true so as long as $a \neq b$ .

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Example Question #4 : Solving Exponential Functions

Solve for , round to the nearest hundredth.

$7(5^{y+1})=7^y$

Possible Answers:

$y \approx 5.54$

$y \approx -1.01$

$y \approx 0.09$

$y \approx 10.57$

$y \approx .18$

Correct answer:

$y \approx 10.57$

Explanation:

Solve for ,

$7(5^{y+1})=7^y$

Note the rules for logarithms,

$\ln(ab) = \ln(a)+\ln(b)$ (1)

$\ln\left(\frac{a}{b} \right )= \ln(a)-\ln(b)$ (2)

$\ln(a^n)=n\ln(a)$ (3)

There is more than one way to get started, but notice that if we divide out by we can proceed as follows,

$5^{y+1}=7^{y-1}$

Take the natural log of both sides,

$\ln(5^{y+1})=\ln(7^{y-1})$

Use rule (3) to pull out the exponents,

$(y+1)\ln(5)=(y-1)\ln(7)$

Expand with the distributive property,

$y\ln(5)+\ln(5)= y\ln(7)-\ln(7)$

Collect terms with on onside and then factor out .

$y[\ln(5)-ln(7)]=-ln(5)-ln(7)$

Divide out to solve for ,

$y = \frac{-\ln(5)-\ln(7)}{\ln(5)-\ln(7)}$ (4)

The numerator can be simplified using rules (2) and (3). Use rule (3) on the $-\ln(5)$ term in the numerator as follows,

$-\ln(5)= (-1)\ln(5)=\ln(5^{-1})=\ln\left(\frac{1}{5} \right )$

So now the numerator can be written,

$\ln\left(\frac{1}{5} \right ) -\ln(7) = \ln\left(\frac{1/5}{7} \right ) = \ln\left(\frac{1}{35} \right )=-\ln(35)$

Bringing back the denominator in equation (4) gives,

$y=-\frac{ \ln(35) }{\ln(5)-\ln(7)}$

$y \approx 10.57$

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Example Question #5 : Solving Exponential Functions

Solve this equation: $2^{x+4}=512$

Possible Answers:

x=1

x=8

x=6

x=5

x=0

Correct answer:

x=5

Explanation:

To solve $2^{x+4}=512$ , use the one-to-one property of exponential equations.

$2^{x+4}=2^9$

x+4=9

$\boldsymbol{x=5}$

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Example Question #3 : Solving Exponential Functions

Find the value of a:

$\log_3(a) = 5$

Possible Answers:

125

243

Correct answer:

243

Explanation:

For this problem, we can use the fact that the exponent and logarithm are inverses of each other. So if we raise both sides up as exponents:

$3^{\log_3(a)} = 3^5$

The left side of the equation cancels and we get that

a = 3^5 = 243

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College Algebra : Exponential and Logarithmic Functions

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #21 : Exponential And Logarithmic Functions

Example Question #1 : Natural Log

Example Question #22 : Exponential And Logarithmic Functions

Example Question #2 : Natural Log

Example Question #1 : Solving Exponential Functions

Example Question #2 : Solving Exponential Functions

Example Question #3 : Solving Exponential Functions

Example Question #4 : Solving Exponential Functions

Example Question #5 : Solving Exponential Functions

Example Question #3 : Solving Exponential Functions

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