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Algebra II : Solving Logarithms

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Example Question #1 : Solving Logarithms

Solve for :

$5^{x} = \left(\frac{1}{25}\right)^{4x-1}$

Possible Answers:

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

$x = \frac{1}{2}$

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

$x = \frac{-2}{7}$

$x = \frac{-1}{2}$

Correct answer:

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

Explanation:

To solve for , first convert both sides to the same base:

$5^{x} = \frac{1}{25}^{4x-1} = (5^{-2})^{4x-1} = 5^{-8x+2}$

Now, with the same base, the exponents can be set equal to each other:

x= - 8x +2

Solving for gives:

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

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Example Question #2 : Solving Logarithms

Solve the equation:

$\log{_{4}}{x^{2}}+\log{_{4}}{3}=1$

Possible Answers:

$\frac{2\sqrt{3}}{3}, -\frac{2\sqrt{3}}{3}$

$\frac{\sqrt{6}}{3}, -\frac{\sqrt{6}}{3}$

$\frac{\sqrt{3}}{3}, -\frac{\sqrt{3}}{3}$

$\frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2}$

1,-1

Correct answer:

$\frac{2\sqrt{3}}{3}, -\frac{2\sqrt{3}}{3}$

Explanation:

$\log{_{4}}{x^{2}}+\log{_{4}}{3}=1$

$\Rightarrow \log{_{4}}{x^{2}}+\log{_{4}}{3}=\log{_{4}}{4}$

$\Rightarrow log{_{4}}{(x^{2}\times 3)}=\log{_{4}}{4}$

$\Rightarrow 3x^{2}=4$

$\Rightarrow x^{2}=\frac{4}{3}$

$\Rightarrow x=\frac{2\sqrt{3}}{3}, x=-\frac{2\sqrt{3}}{3}$

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Example Question #3 : Solving Logarithms

Use $\small \small \log_23\approx 1.5850$ to approximate the value of $\small \log_2 24$ .

Possible Answers:

$\small \log_2 24\approx 3.4150$

$\small \log_2 24\approx 12.5850$

$\small \log_2 24\approx 3.5850$

$\small \log_2 24\approx 4.5850$

$\small \log_2 24\approx 12.6800$

Correct answer:

$\small \log_2 24\approx 4.5850$

Explanation:

Rewrite as a product that includes the number :

$\small \log_2 24 = \log_2 (2^3 \cdot 3)$

Then we can split up the logarithm using the Product Property of Logarithms:

$\small \log_2 (2^3\cdot 3) = \log_2 2^3 + \log_2 3$

$\small =3+\log_2 3$

$\small \approx 3 + 1.5850$

Thus,

$\small \log_2 24 \approx 4.5850$ .

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Example Question #4 : Solving Logarithms

Solve for .

$log_{9}^{ }(2x+1)=2$

Possible Answers:

x=40

x=20

x=41

$x=\frac{17}{2}$

$x=\frac{81}{2}$

Correct answer:

x=40

Explanation:

Rewrite in exponential form:

$9^{2}=2x+1$

Solve for x:

81=2x+1

x=40

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Example Question #5 : Solving Logarithms

Solve the following equation:

ln(3x+8)=0

Possible Answers:

$-\frac{7}{3}$

$\frac{8}{7}$

$\frac{1}{3}$

$-\frac{8}{3}$

Correct answer:

$-\frac{7}{3}$

Explanation:

For this problem it is helpful to remember that,

ln(3x+8) = 0 is equivalent to ln(3x+8) = ln(1) because ln(1)=0

Therefore we can set what is inside of the parentheses equal to each other and solve for as follows:

3x+8=1

3x=-7

$x=-\frac{7}{3}$

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Example Question #6 : Solving Logarithms

Solve for :

log_6x=5

Possible Answers:

10000

15625

7776

Correct answer:

7776

Explanation:

To solve this logarithm, we need to know how to read a logarithm. A logarithm is the inverse of an exponential function. If a exponential equation is

a^c = b

then its inverse function, or logarithm, is

log_ab=c

Therefore, for this problem, in order to solve for , we simply need to solve

6^5=x

which is 7776 .

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Example Question #7 : Solving Logarithms

$\log x+\log\left ( x+21\right )=2$

Solve for .

Possible Answers:

100

Correct answer:

Explanation:

Logs are exponential functions using base 10 and a property is that you can combine added logs by multiplying.

$x(x+21)=10^{2}$

$x^{2}+21x=100$

(x+25)(x-4)=0

You cannot take the log of a negative number. x=-25 is extraneous.

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Example Question #151 : Logarithms

If $\log_{x}36=2$ , which of the following is a possible value for ?

Possible Answers:

Correct answer:

Explanation:

This question is testing the definition of logs. $\log_{a}b=x$ is the same as $a^{x}=b$ .

In this case, $\log_{x}36=2$ can be rewritten as $x^{2}=36$ .

Taking square roots of both sides, you get $x=\pm 6$ . Since only the positive answer is one of the answer choices, is the correct answer.

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Example Question #9 : Solving Logarithms

Rewriting Logarithms in Exponential Form

Solve for below:

$3^{x}=15$

Which of the below represents this function in log form?

Possible Answers:

$log{_{x}}{3}=15$

$log{_{3}}{15}=x$

$log{_{x}}{15}=3$

$log{_{15}}{3}=x$

Correct answer:

$log{_{3}}{15}=x$

Explanation:

The first step is to rewrite this equation in log form.

When rewriting an exponential function as a log we must remember that the form of an exponential is:

$b^{^{x}}=a$

When this is rewritten in log form it is:

$log{_{b}}(a)=x$ .

Therefore we have $3^{x}=15$ which when rewritten gives us,

$log{_{3}}{15}=x$ .

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Example Question #10 : Solving Logarithms

Solve for :

$log_3x=\frac{1}{3}log_364+\frac{1}{2}log_349$ .

Possible Answers:

Not enough information

x=45.8

x=28

x=11

x=522.7

Correct answer:

x=28

Explanation:

Use the rule of Exponents of Logarithms to turn all the multipliers into exponents:

$log_3x=log_364^\frac{1}{3}+log_349^\frac{1}{2}$ .

Simplify by applying the exponents: log_3x=log_34+log_37 .

According to the law for adding logarithms, log_b(mn) = log_b(m) + log_b(n) .

Therefore, multiply the 4 and 7.

log_3x=log_328 .

Because both sides have the same base, x=28 .

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Algebra II : Solving Logarithms

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1 : Solving Logarithms

Example Question #2 : Solving Logarithms

Example Question #3 : Solving Logarithms

Example Question #4 : Solving Logarithms

Example Question #5 : Solving Logarithms

Example Question #6 : Solving Logarithms

Example Question #7 : Solving Logarithms

Example Question #151 : Logarithms

Example Question #9 : Solving Logarithms

Example Question #10 : Solving Logarithms

All Algebra II Resources

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