Solving Logarithmic Functions - College Algebra

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Question

Solve for :

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

Answer

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:

Solving for gives:

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

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Question

Solve the equation:

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

Answer

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

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

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

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

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

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

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Question

Solve for .

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

Answer

Rewrite in exponential form:

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

Solve for x:

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Question

Solve the following equation:

Answer

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:

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

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Question

Solve this logarithmic equation: $\log_{10}(5x-2)=\log_{10}(x+6)$

Answer

To solve this problem you must be familiar with the one-to-one logarithmic property.

$\log_{10}x=\log_{10}y$ if and only if x=y. This allows us to eliminate to logarithmic functions assuming they have the same base.

$\log_{10}(5x-2)=\log_{10}(x+6)$

one-to-one property:

isolate x's to one side:

move constant:

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Question

$\log_{5}x=3$

Answer

To solve this equation, remember log rules

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

This rule can be applied here so that

and

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Question

Solve the equation:

$5e^{3x+4}-10=0$

Answer

Get all the terms with e on one side of the equation and constants on the other.

$5e^{3x+4}-10=0$

$e^{3x+4}=2$

Apply the logarithmic function to both sides of the equation.

$ln(e^{3x+4})=ln2$

$x=\frac{-4+ln2}{3}$

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Question

Solve the equation:

Answer

Recall the rules of logs to solve this problem.

First, when there is a coefficient in front of log, this is the same as log with the inside term raised to the outside coefficient.

$log(x^{2})+log(5)=3$

Also, when logs of the same base are added together, that is the same as the two inside terms multiplied together.

In mathematical terms:

$log(a)+log(b)=log(a\cdot b)$

Thus our equation becomes,

$log(5x^{2})=3$

To simplify further use the rule,

$log_b(x)\rightarrow b^{log_b(x)}=x$

$10^{log(5x^{2})}=10^{3}$

$5x^{2}=1000$

$x^{2}=200$

$x=\pm10\sqrt{2}$ .

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Question

Solve the following equation for t

Answer

Solve the following equation for t

We can solve this equation by rewriting it as an exponential equation:

Next, take the fourth root to get:

$t=\sqrt[4]{625}=5$

We can check our work vie the following:

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Question

Solve for ,

$\ln(x^2 +1)=3$

Answer

$\ln(x^2+1)=3$

The first step step is to carry out the inverse operation of the natural logarithm,

$e^{\ln(x^2+1)}=e^3$

We can use the property $e^{ln(u)}=u$ to simplify the left side of the equation to obtain,

Solve for ,

$x = \pm \sqrt{e^3-1}$

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Question

Solve this equation: $e^{2x}-3e^x+2=0$

Answer

$e^{2x}-3e^x+2=0$

Note that this equation is a quadratic because $e^{2x}=(e^x)^2$

Factor:

Set each factor equal to 0 and solve:

$e^x-2=0\rightarrow e^x=2 \rightarrow \boldsymbol{x=ln\hspace{1mm} 2}$

$e^x-1=0\rightarrow e^x=1 \rightarrow \boldsymbol{x=0}$

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Question

Solve this logarithmic equation:

Answer

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)

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Question

Solve for x:

$log_{3}x=2$

Answer

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:

$log_{3}x=2$

$3^{2}=x$

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Question

Solve for x:

$log_{8}x=6$

Answer

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:

$log_{8}x=6$

$8^{6}=x$

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Question

Solve for x:

$log_{5}x=3$

Answer

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:

$log_{5}x=3$

$5^{3}=x$

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Question

Solve for x:

$log_{2}x=8$

Answer

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:

$log_{2}x=8$

$2^{8}=x$

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Question

Solve for x:

$log_{10}x=4$

Answer

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:

$log_{10}x=4$

$10^{4}=x$

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Question

Solve for x:

$log_{14}x=2$

Answer

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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Question

Solve for x:

$log_{6}x=3$

Answer

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:

$log_{6}x=3$

$6^{3}=x$

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Question

Solve for x:

$log_{6}x=4$

Answer

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:

$log_{6}x=4$

$6^{4}=x$

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