Solving Logarithmic Functions

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College Algebra › Solving Logarithmic Functions

Questions 1 - 10

$\log_{5}x=3$

$\frac{1}{25}$

$\frac{1}{125}$

Explanation

To solve this equation, remember log rules

$\log_{b}x=y; b^{y}=x$ .

This rule can be applied here so that

and

Solve for x:

$log_{2}x=8$

Explanation

In order to solve for x in a logarithmic function, we need to change it into exponential form. That looks as follows:

$log_{b}x=y\rightarrow b^{y}=x$

For this problem, manipulate the log and solve:

$log_{2}x=8$

$2^{8}=x$

Solve for x:

$log_{10}x=4$

Explanation

In order to solve for x in a logarithmic function, we need to change it into exponential form. That looks as follows:

$log_{b}x=y\rightarrow b^{y}=x$

For this problem, manipulate the log and solve:

$log_{10}x=4$

$10^{4}=x$

Solve for .

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

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

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

Explanation

Rewrite in exponential form:

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

Solve for x:

What is the correct value of ?

$\frac{1}{1000}$

Explanation

Divide by three on both sides.

$\frac{3log_b (10)}{3}=\frac{1}{3}$

$log_b (10)=\frac{1}{3}$

If we would recall and $log_{10}(100) = 2$ , this indicates that:

$b^{\frac{1}{3}} = 10$

Cube both sides to isolate b.

$(b^{\frac{1}{3}}) ^3= 10^3$

The answer is:

Solve for x:

$log_{6}x=4$

Explanation

In order to solve for x in a logarithmic function, we need to change it into exponential form. That looks as follows:

$log_{b}x=y\rightarrow b^{y}=x$

For this problem, manipulate the log and solve:

$log_{6}x=4$

$6^{4}=x$

Solve this logarithmic equation:

None of these.

Explanation

Exponentiate:

$e^{ln2x-2}=e^5$

Add two to both sides:

$2x-2=e^5\rightarrow 2x=e^5+2$

Divide both sides by 2:

$x=\frac{e^5}{2}+1=\boldsymbol{75.2}$ (rounded answer)

Solve the following equation:

$-\frac{7}{3}$

$-\frac{8}{3}$

$\frac{1}{3}$

$\frac{8}{7}$

Explanation

For this problem it is helpful to remember that,

is equivalent to because

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

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

Solve for x:

$log_{6}x=3$

Explanation

In order to solve for x in a logarithmic function, we need to change it into exponential form. That looks as follows:

$log_{b}x=y\rightarrow b^{y}=x$

For this problem, manipulate the log and solve:

$log_{6}x=3$

$6^{3}=x$

Solve for x:

$log_{14}x=2$

Explanation

In order to solve for x in a logarithmic function, we need to change it into exponential form. That looks as follows:

$log_{b}x=y\rightarrow b^{y}=x$

For this problem, manipulate the log and solve:

$log_{14}x=2$

$14^{2}=x$

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