Solve for x 3 log base 6 of 2+2 log base 6 of 3- log base 6 of x=2

$3 \log_{6} (2) + 2 \log_{6} (3) - \log_{6} (x) = 2$

Simplify the left side.

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Simplify each term.

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Simplify $3 \log_{6} (2)$ by moving $3$ inside the logarithm.

$\log_{6} (2^{3}) + 2 \log_{6} (3) - \log_{6} (x) = 2$

Raise $2$ to the power of $3$ .

$\log_{6} (8) + 2 \log_{6} (3) - \log_{6} (x) = 2$

Simplify $2 \log_{6} (3)$ by moving $2$ inside the logarithm.

$\log_{6} (8) + \log_{6} (3^{2}) - \log_{6} (x) = 2$

Raise $3$ to the power of $2$ .

$\log_{6} (8) + \log_{6} (9) - \log_{6} (x) = 2$

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

$\log_{6} (8 \cdot 9) - \log_{6} (x) = 2$

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

$\log_{6} (\frac{8 \cdot 9}{x}) = 2$

Multiply $8$ by $9$ .

$\log_{6} (\frac{72}{x}) = 2$

Rewrite $\log_{6} (\frac{72}{x}) = 2$ 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 $b^{y} = x$ .

$6^{2} = \frac{72}{x}$

Solve for $x$

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

$36 = \frac{72}{x}$

Rewrite the equation as $\frac{72}{x} = 36$ .

$\frac{72}{x} = 36$

Solve for $x$ .

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Multiply each term by $x$ and simplify.

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Multiply each term in $\frac{72}{x} = 36$ by $x$ .

$\frac{72}{x} \cdot x = 36 \cdot x$

Cancel the common factor of $x$ .

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Cancel the common factor.

$\frac{72}{x} \cdot x = 36 \cdot x$

Rewrite the expression.

$72 = 36 \cdot x$

$72 = 36 x$

Rewrite the equation as $36 x = 72$ .

$36 x = 72$

Divide each term by $36$ and simplify.

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Divide each term in $36 x = 72$ by $36$ .

$\frac{36 x}{36} = \frac{72}{36}$

Cancel the common factor of $36$ .

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Cancel the common factor.

$\frac{36 x}{36} = \frac{72}{36}$

Divide $x$ by $1$ .

$x = \frac{72}{36}$

Divide $72$ by $36$ .

$x = 2$

Solve for x 3 log base 6 of 2+2 log base 6 of 3- log base 6 of x=2

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