Solve for x log base 6 of 9x- log base 6 of 4=2

$\log_{6} (9 x) - \log_{6} (4) = 2$

Step 1

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

$\log_{6} (\frac{9 x}{4}) = 2$

Step 2

Rewrite $\log_{6} (\frac{9 x}{4}) = 2$ 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 $b^{y} = x$ .

$6^{2} = \frac{9 x}{4}$

Step 3

Solve for $x$ .

Rewrite the equation as $\frac{9 x}{4} = 6^{2}$ .

$\frac{9 x}{4} = 6^{2}$

Multiply both sides of the equation by $\frac{4}{9}$ .

$\frac{4}{9} \cdot \frac{9 x}{4} = \frac{4}{9} \cdot 6^{2}$

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{4}{9} \cdot \frac{9 x}{4}$ .

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4}{9} \cdot \frac{9 x}{4} = \frac{4}{9} \cdot 6^{2}$

Rewrite the expression.

$\frac{1}{9} (9 x) = \frac{4}{9} \cdot 6^{2}$

Cancel the common factor of $9$ .

Factor $9$ out of $9 x$ .

$\frac{1}{9} (9 (x)) = \frac{4}{9} \cdot 6^{2}$

Cancel the common factor.

$\frac{1}{9} (9 x) = \frac{4}{9} \cdot 6^{2}$

Rewrite the expression.

$x = \frac{4}{9} \cdot 6^{2}$

Simplify the right side.

Simplify $\frac{4}{9} \cdot 6^{2}$ .

Raise $6$ to the power of $2$ .

$x = \frac{4}{9} \cdot 36$

Cancel the common factor of $9$ .

Factor $9$ out of $36$ .

$x = \frac{4}{9} \cdot (9 (4))$

Cancel the common factor.

$x = \frac{4}{9} \cdot (9 \cdot 4)$

Rewrite the expression.

$x = 4 \cdot 4$

Multiply $4$ by $4$ .

$x = 16$

Solve for x log base 6 of 9x- log base 6 of 4=2

Related Questions

Support LCMGCF.com by disabling your adblocker.