Finite Math Examples

$\log_{g} (x - 12) + \log_{g} (x) = 2$

Step 1

Solve the equation for $g$ .

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Step 1.1

Simplify the left side.

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Step 1.1.1

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log_{g} ((x - 12) x) = 2$

Step 1.1.2

Apply the distributive property.

$\log_{g} (x \cdot x - 12 x) = 2$

Step 1.1.3

Multiply $x$ by $x$ .

$\log_{g} (x^{2} - 12 x) = 2$

Step 1.2

Rewrite $\log_{g} (x^{2} - 12 x) = 2$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$g^{2} = x^{2} - 12 x$

Step 1.3

Solve for $g$ .

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Step 1.3.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$g = \pm \sqrt{x^{2} - 12 x}$

Step 1.3.2

Factor $x$ out of $x^{2} - 12 x$ .

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Step 1.3.2.1

Factor $x$ out of $x^{2}$ .

$g = \pm \sqrt{x \cdot x - 12 x}$

Step 1.3.2.2

Factor $x$ out of $- 12 x$ .

$g = \pm \sqrt{x \cdot x + x \cdot - 12}$

Step 1.3.2.3

Factor $x$ out of $x \cdot x + x \cdot - 12$ .

$g = \pm \sqrt{x (x - 12)}$

Step 1.3.3

The complete solution is the result of both the positive and negative portions of the solution.

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Step 1.3.3.1

First, use the positive value of the $\pm$ to find the first solution.

$g = \sqrt{x (x - 12)}$

Step 1.3.3.2

Next, use the negative value of the $\pm$ to find the second solution.

$g = - \sqrt{x (x - 12)}$

Step 1.3.3.3

The complete solution is the result of both the positive and negative portions of the solution.

$g = \sqrt{x (x - 12)}$

$g = - \sqrt{x (x - 12)}$

$g = \sqrt{x (x - 12)}$

$g = - \sqrt{x (x - 12)}$

$g = \sqrt{x (x - 12)}$

$g = - \sqrt{x (x - 12)}$

$g = \sqrt{x (x - 12)}$

$g = - \sqrt{x (x - 12)}$

Step 2

A linear equation is an equation of a straight line, which means that the degree of a linear equation must be $0$ or $1$ for each of its variables. In this case, the degree of the variable in the equation violates the linear equation definition, which means that the equation is not a linear equation.

Not Linear