Solve for x log base 4 of (x-1)^3=9

$\log_{4} ({(x - 1)}^{3}) = 9$

Rewrite $\log_{4} ({(x - 1)}^{3}) = 9$ 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 $b^{y} = x$ .

$4^{9} = {(x - 1)}^{3}$

Solve for $x$ .

Rewrite the equation as ${(x - 1)}^{3} = 4^{9}$ .

${(x - 1)}^{3} = 4^{9}$

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

$x - 1 = \sqrt[3]{4^{9}}$

Simplify $\sqrt[3]{4^{9}}$ .

Raise $4$ to the power of $9$ .

$x - 1 = \sqrt[3]{262144}$

Rewrite $262144$ as $64^{3}$ .

$x - 1 = \sqrt[3]{64^{3}}$

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

$x - 1 = 64$

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

Add $1$ to both sides of the equation.

$x = 64 + 1$

Add $64$ and $1$ .

$x = 65$

Solve for x log base 4 of (x-1)^3=9

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