Solve for x 8^(x-2) = square root of 8

$8^{x - 2} = \sqrt{8}$

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{8}$ as $8^{\frac{1}{2}}$ .

$8^{x - 2} = 8^{\frac{1}{2}}$

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$x - 2 = \frac{1}{2}$

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

Add $2$ to both sides of the equation.

$x = \frac{1}{2} + 2$

To write $2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = \frac{1}{2} + 2 \cdot \frac{2}{2}$

Combine $2$ and $\frac{2}{2}$ .

$x = \frac{1}{2} + \frac{2 \cdot 2}{2}$

Combine the numerators over the common denominator.

$x = \frac{1 + 2 \cdot 2}{2}$

Simplify the numerator.

Multiply $2$ by $2$ .

$x = \frac{1 + 4}{2}$

Add $1$ and $4$ .

$x = \frac{5}{2}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{5}{2}$

Decimal Form:

$x = 2.5$

Mixed Number Form:

$x = 2 \frac{1}{2}$