Solve for x log base 2 of x-1=3

$\log_{2} (x - 1) = 3$

Rewrite $\log_{2} (x - 1) = 3$ 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 $b^{y} = x$ .

$2^{3} = x - 1$

Solve for $x$ .

Rewrite the equation as $x - 1 = 2^{3}$ .

$x - 1 = 2^{3}$

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

$x - 1 = 8$

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

Add $1$ to both sides of the equation.

$x = 8 + 1$

Add $8$ and $1$ .

$x = 9$

Solve for x log base 2 of x-1=3

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