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Precalculus : Solve Logarithmic Equations

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Example Question #1 : Solve Logarithmic Equations

Evaluate a logarithm.

What is $log_{10}1000$ ?

Possible Answers:

$\frac{1}{3}$

100

Correct answer:

Explanation:

The derifintion of logarithm is:

$log_{a}y=x \Leftrightarrow a^{x}=y.$

In this problem,

a= 10, y=1000

$10^{3}=1000$

Therefore, x=3

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Example Question #2 : Solve Logarithmic Equations

Solve for in the following logarithmic equation:

$log_{10} x^2 - log_{10} 2x = 2$

Possible Answers:

None of the other choices

x = 0

x = 1000

x=100

x = 10

Correct answer:

None of the other choices

Explanation:

Using the rules of logarithms,

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

Hence,

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

So exponentiate both sides with a base 10:

$10^{log_{10} \frac{x}{2}} = 10^{2}$

The exponent and the logarithm cancel out, leaving:

$\frac{x}{2}=10^2$

$x=100\cdot 2=200$

This answer does not match any of the answer choices, therefore the answer is 'None of the other choices'.

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Example Question #1 : Solve Logarithmic Equations

Solve the following logarithmic equation:

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

Possible Answers:

x=-4

x=-1

x=2

x=4

x=1

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 #1162 : Pre Calculus

Solve the equation for .

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

Possible Answers:

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

None of the other answers.

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

ab-e^c

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

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 #1163 : Pre Calculus

Solve for :

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

Possible Answers:

x=-3

x=54

x=3

x=3.67

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

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 #1164 : Pre Calculus

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

Solve for .

Possible Answers:

$e^{3/2}$

$e^{-3/2}$

$e^{4/3}$

$e^{-1/2}$

$e^{2/3}$

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 #1165 : Pre Calculus

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

Possible Answers:

x=e^2-2

x=e^4-4

x=e^4-2

None of the other answers.

x=e^2+4

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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Precalculus : Solve Logarithmic Equations

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Solve Logarithmic Equations

Example Question #2 : Solve Logarithmic Equations

Example Question #1 : Solve Logarithmic Equations

Example Question #1162 : Pre Calculus

Example Question #1163 : Pre Calculus

Example Question #1164 : Pre Calculus

Example Question #1165 : Pre Calculus

Example Question #1166 : Pre Calculus

Example Question #1167 : Pre Calculus

Example Question #1168 : Pre Calculus

All Precalculus Resources

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