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Precalculus : Pre-Calculus

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12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3 : Solve Logarithmic Equations

Solve the following logarithmic equation:

$2\log _3(x)=\log _3(4)+\log _3(x-1)$

Possible Answers:

x=1

x=-1

x=-4

x=4

x=2

Correct answer:

x=2

Explanation:

In order to solve this equation, we must apply several properties of logarithms. First we notice the term on the left side of the equation, which we can rewrite using the following property:

$a\log_b (x)=\log_b (x^a)$

Where a is the coefficient of the logarithm and b is some arbitrary base. Next we look at the right side of the equation, which we can rewrite using the following property for the addition of logarithms:

$\log _ba+\log _bc=\log_b (a\cdot c)$

Using both of these properties, we can rewrite the logarithmic equation as follows:

$2\log _3(x)=\log _3(4)+\log _3(x-1)$

$\log _3(x^2)=\log _3(4(x-1))$

We have the same value for the base of the logarithm on each side, so the equation then simplifies to the following:

$x^2=4x-4\rightarrow x^2-4x+4=0$

Which we can then factor to solve for :

$(x-2)(x-2)=0\rightarrow x=2$

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Example Question #4 : Solve Logarithmic Equations

Solve the equation for .

$\ln\frac{cx}{ab+c} = b+c$

Possible Answers:

$x = \frac{e^a}{e^{bc}}}(ab+c)$

$x = \frac{e^b}{c(ab+c)e^c}$

ab-e^c

$x = \frac{e^be^c(ab+c)}{c}}$

None of the other answers.

Correct answer:

$x = \frac{e^be^c(ab+c)}{c}}$

Explanation:

We solve the equation as follows:

$\ln \frac{cx}{ab+c} = b+c$

Exponentiate both sides.

$\frac{cx}{ab+c}= e^{b+c}$

Apply the power rule on the right hand side.

$\frac{cx}{ab+c} = e^be^c$

Multiply by ab+c .

cx = e^be^c(ab+c)

Divide by .

$x =\frac{e^be^c(ab+c)}{c}$

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Example Question #4 : Solve Logarithmic Equations

Solve for :

log(27)- 2log(x)=log(x)

Possible Answers:

x=54

x=3

x=-3

$x=-\frac{1}{27}$

x=3.67

Correct answer:

x=3

Explanation:

First, simplify the logarithmic expressions on the left side of the equation:

2log(x) can be re-written as log(x^2) .

Now we have:

log(27)-log(x^2)=log(x) .

The left can be consolidated into one log expression using the subtraction rule:

$log\left(\frac{27}{x^2}\right)=log(x)$ .

We now have log on both sides, so we can be confident that whatever is inside these functions is equal:

$\frac{27}{x^2}=x$ to continue solving, multiply by x^2 on both sides:

27 = x^3 take the cube root:

3 = x

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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 #11 : Solve Logarithmic Equations

Solve the following logarithmic equation:

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

Possible Answers:

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

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

x=5

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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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #3 : Solve Logarithmic Equations

Example Question #4 : Solve Logarithmic Equations

Example Question #4 : Solve Logarithmic Equations

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 #11 : Solve Logarithmic Equations

Example Question #71 : Exponential And Logarithmic Functions

All Precalculus Resources

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