Solve for x log base 4 of (x-6)^3=6

$\log_{4} ({(x - 6)}^{3}) = 6$

Step 1

Rewrite $\log_{4} ({(x - 6)}^{3}) = 6$ 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 $b^{y} = x$ .

$4^{6} = {(x - 6)}^{3}$

Step 2

Solve for $x$ .

Rewrite the equation as ${(x - 6)}^{3} = 4^{6}$ .

${(x - 6)}^{3} = 4^{6}$

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

$x - 6 = \sqrt[3]{4^{6}}$

Simplify $\sqrt[3]{4^{6}}$ .

Raise $4$ to the power of $6$ .

$x - 6 = \sqrt[3]{4096}$

Rewrite $4096$ as $16^{3}$ .

$x - 6 = \sqrt[3]{16^{3}}$

Pull terms out from under the radical, assuming real numbers.

$x - 6 = 16$

Move all terms not containing $x$ to the right side of the equation.

Add $6$ to both sides of the equation.

$x = 16 + 6$

Add $16$ and $6$ .

$x = 22$

Solve for x log base 4 of (x-6)^3=6

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