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Precalculus : Exponential and Logarithmic Functions

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Example Questions

Example Question #1 : Solve Logarithmic Equations

$(\ln (x^2))^2=\ln (x^4)+3$ .

Solve for .

Possible Answers:

$e^{4/3}$

$e^{-3/2}$

$e^{2/3}$

$e^{-1/2}$

$e^{3/2}$

Correct answer:

$e^{3/2}$

Explanation:

First bring the inside exponent in front of the natural log.

$(\ln (x^2))^2=\ln (x^4)+3\rightarrow(2\ln (x))^2=4\ln (x)+3$ .

Next simplify the first term and bring all the terms on one side of the equation.

$4(\ln (x))^2-4\ln (x)-3=0$ .

Next, let set

$\ln (x)=A$ , so 4A^2-4A-3=0 .

Now use the quadratic formula to solve for .

$\\ \frac{-b\pm \sqrt{b^2-4(a)(c)}}{2a}\\ \\=\frac{4\pm\sqrt{16-4(4)(-3)}}{2(4)}\\ \\=\frac{4\pm \sqrt{16+48}}{8}\\ \\=4\pm \frac{\sqrt{64}}{8}\\ \\=\frac{4\pm 8}{8}\\ \\=\frac{3}{2}; -\frac{1}{2}$

and thus, $A=-\frac{1}{2}$ and $A=\frac{3}{2}$ .

Now substitute with $\ln (x)$ .

So, $\ln (x)=-\frac{1}{2}\rightarrow \O$ since $x\nleqslant0$ and $\ln (x)=\frac{3}{2}\rightarrow x=e^{3/2}$ .

Thus, $x=e^{3/2}$ .

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Example Question #6 : Solve Logarithmic Equations

Solve the logarithmic equation: $\ln\sqrt{x+2}=2$

Possible Answers:

x=e^4-4

x=e^2+4

None of the other answers.

x=e^2-2

x=e^4-2

Correct answer:

x=e^4-2

Explanation:

$\ln\sqrt{x+2}=2$

Exponentiate each side to cancel the natural log:

$e^{\ln\sqrt{x+2}}=e^2$

$\sqrt{x+2}=e^2$

Square both sides:

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

x+2=e^4

Isolate x:

x=e^4-2

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Example Question #1166 : Pre Calculus

Solve for x: $\log(2x+1) = 3$

Possible Answers:

499.5

9.5

Correct answer:

499.5

Explanation:

The base of a logarithm is 10 by default:

$\log _{10 } (2 x + 1 ) = 3$ convert to exponent to isolate x

10^3 = 2x + 1

1,000 = 2x + 1 subtract 1 from both sides

999 = 2x divide both sides by 2

499.5 = x

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Example Question #1167 : Pre Calculus

Solve for x: $\log _3 x - \log _3 7 = 2$

Possible Answers:

Correct answer:

Explanation:

First, condense the left side into one logarithm:

$\log _3 \frac{x}{7 } = 2$ convert to an exponent

$3^2 = \frac{x}{7 }$

$9 = \frac{x}{7}$ multiply both sides by 7

63 = x

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Example Question #1168 : Pre Calculus

Solve for x: $\log _4 x + \log _ 4 (3x - 7 ) = 3$

Possible Answers:

3.95, - 2.40

5.93 , -3.60

no solution

2.5

8.90, - 5.40

Correct answer:

5.93 , -3.60

Explanation:

First, consolidate the left side into one logarithm:

$\log _4 (3x^2 - 7x ) = 3$ convert to an exponent

3x^2 - 7x = 64 subtract 64 from both sides

3x^2 - 7 x - 64 = 0 now we can solve using the quadratic formula:

$x = \frac{7 \pm \sqrt{49 - 4(3)(-64)}}{6 } \approx 5.93, -3.60$

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Example Question #1169 : Pre Calculus

Solve the following logarithmic equation:

2log(x)=log(2)+log(x+3)

Possible Answers:

x=5

$x=2\pm \sqrt{28}$

$x=1\pm \sqrt{7}$

x=-3

$x=\sqrt{7}$

Correct answer:

$x=1\pm \sqrt{7}$

Explanation:

We must first use some properties of logs to rewrite the equation. First, using the power rule, which says

$b\cdot log(x)=log(b^x)$

we can rewrie the left side of the equation, as below:

log(x^2)=log(2)+log(x+3)

Now, we want to use the product property of logarithms to condense the right side into just one log, as below:

log(x^2)=log(2(x+3))

Because the logs are both base 10, we can simply set the insides equal, like this:

x^2=2(x+3)

Now we have a polynomial to solve.

x^2=2x+6

x^2-2x-6=0

Using the quadratic formula to solve for x

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

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Example Question #71 : Exponential And Logarithmic Functions

Solve this logarithm for :

$25e^{0.02x}=100$

Possible Answers:

x=75.4

x=86

x=69.3

None of these.

x=89.4

Correct answer:

x=69.3

Explanation:

$25e^{0.02x}=100$

Divide both sides by 25 to isolate the exponential function:

$e^{0.02x}=4$

Take the natural log of both sides:

$\ln (e^{0.02x})=\ln(4)$

Solve for x:

$0.02x=\ln 4$

$x=\frac{ln4}{0.02}=\mathbf{69.3}$

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Example Question #71 : Exponential And Logarithmic Functions

Solve the following logarithmic equation.

$log_{2}\,x +log_{2}\,(4-x)=2$

Possible Answers:

x=0

x=2

x=1

x=-4

Correct answer:

x=2

Explanation:

In order to solve the logarithmic equation, we use the following property

$log\,a+log\,b=log\,ab$

As such

$log_{2}\,x +log_{2}\,(4-x)=log_{2}\,x(4-x)=log_2\,4x-x^2=2$

And converting from logarithmic form to exponential form

$log_a\,c=b \Rightarrow a^b=c$

we get

2^2=4x-x^2

Solving for x

x^2-4x+4=0

And because the square of a difference is given as this equation through factoring

(a-b)^2=a^2-2ab+b^2

we have

(x-2)^2=0

which implies

x=2

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Example Question #11 : Solve Logarithmic Equations

Express the log in its expanded form:

$\log\bigg(\frac{x^3y}{z}\bigg)$

Possible Answers:

None of the other answers

$\log^{3}(x)+\log(y)-\log(z)$

$3\log(x)+\log(y)- \log(z)$

$3\log(x)+\log(y)+\log(z)$

$\log(x)+\log(y)-\log(z)$

Correct answer:

$3\log(x)+\log(y)- \log(z)$

Explanation:

You need to know the Laws of Logarithms in order to solve this problem. The ones specifically used in this problem are the following:

$\log(xy) = \log (x)+\log(y)$

$\log\bigg(\frac{x}{y}\bigg) = \log(x)-\log(y)$

$\log(x^y) = y\cdot \log(x)$

Let's take this one variable at a time starting with expanding z:

$\log\bigg(\frac{x^3y}{z}\bigg) = \log(x^3y)-\log(z)$

Now y:

$\log(x^3y)-\log(z) = \log(x^3)+\log(y)-\log(z)$

And finally expand x:

$\log(x^3)+\log(y)-\log(z) = 3\log(x)+\log(y)-\log(z)$

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Example Question #71 : Exponential And Logarithmic Functions

What is ln(2x)-ln(2) equivalent to?

Possible Answers:

$\ln(2)$

$\ln(x)$

$\frac{\ln(2x)}{2}$

$\ln (2x-2)$

Correct answer:

$\ln(x)$

Explanation:

Using the properties of logarithms,

$\log(a)-\log(b)=\log \bigg( \frac{a}{b}\bigg)$

the expression can be rewritten as

$\ln(2x)-\ln(2)=\ln\bigg(\frac{2x}{2}\bigg)$

which simplifies to $\ln(x)$ .

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Precalculus : Exponential and Logarithmic Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Solve Logarithmic Equations

Example Question #6 : Solve Logarithmic Equations

Example Question #1166 : Pre Calculus

Example Question #1167 : Pre Calculus

Example Question #1168 : Pre Calculus

Example Question #1169 : Pre Calculus

Example Question #71 : Exponential And Logarithmic Functions

Example Question #71 : Exponential And Logarithmic Functions

Example Question #11 : Solve Logarithmic Equations

Example Question #71 : Exponential And Logarithmic Functions

All Precalculus Resources

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