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

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

Step 1

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

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

Step 2

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

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

Step 3

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

Add $2$ to both sides of the equation.

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $2$ by $2$ .

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

Add $1$ and $4$ .

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

Step 4

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$n = 2.5$

Mixed Number Form:

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