Finite Math Examples

$\log_{5} (x) + \log_{5} (3) = \log_{5} (6)$

Step 1

Subtract $\log_{5} (6)$ from both sides of the equation.

$\log_{5} (x) + \log_{5} (3) - \log_{5} (6) = 0$

Step 2

Simplify $\log_{5} (x) + \log_{5} (3) - \log_{5} (6)$ .

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Step 2.1

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

$\log_{5} (x \cdot 3) - \log_{5} (6) = 0$

Step 2.2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{5} (\frac{x \cdot 3}{6}) = 0$

Step 2.3

Cancel the common factor of $3$ and $6$ .

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Step 2.3.1

Factor $3$ out of $x \cdot 3$ .

$\log_{5} (\frac{3 \cdot x}{6}) = 0$

Step 2.3.2

Cancel the common factors.

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Step 2.3.2.1

Factor $3$ out of $6$ .

$\log_{5} (\frac{3 \cdot x}{3 \cdot 2}) = 0$

Step 2.3.2.2

Cancel the common factor.

$\log_{5} (\frac{3 \cdot x}{3 \cdot 2}) = 0$

Step 2.3.2.3

Rewrite the expression.

$\log_{5} (\frac{x}{2}) = 0$

Step 3

Rewrite $\log_{5} (\frac{x}{2}) = 0$ 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 $b^{y} = x$ .

$5^{0} = \frac{x}{2}$

Step 4

Solve for $x$ .

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Step 4.1

Rewrite the equation as $\frac{x}{2} = 5^{0}$ .

$\frac{x}{2} = 5^{0}$

Step 4.2

Multiply both sides of the equation by $2$ .

$2 \frac{x}{2} = 2 \cdot 5^{0}$

Step 4.3

Simplify both sides of the equation.

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Step 4.3.1

Simplify the left side.

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Step 4.3.1.1

Cancel the common factor of $2$ .

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Step 4.3.1.1.1

Cancel the common factor.

$2 \frac{x}{2} = 2 \cdot 5^{0}$

Step 4.3.1.1.2

Rewrite the expression.

$x = 2 \cdot 5^{0}$

Step 4.3.2

Simplify the right side.

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Step 4.3.2.1

Simplify $2 \cdot 5^{0}$ .

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Step 4.3.2.1.1

Anything raised to $0$ is $1$ .

$x = 2 \cdot 1$

Step 4.3.2.1.2

Multiply $2$ by $1$ .

$x = 2$