Solve for x 7^(x+3) = square root of 7

$7^{x + 3} = \sqrt{7}$

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{7}$ as $7^{\frac{1}{2}}$ .

$7^{x + 3} = 7^{\frac{1}{2}}$

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

$x + 3 = \frac{1}{2}$

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

Subtract $3$ from both sides of the equation.

$x = \frac{1}{2} - 3$

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = \frac{1}{2} - 3 \cdot \frac{2}{2}$

Combine $- 3$ and $\frac{2}{2}$ .

$x = \frac{1}{2} + \frac{- 3 \cdot 2}{2}$

Combine the numerators over the common denominator.

$x = \frac{1 - 3 \cdot 2}{2}$

Simplify the numerator.

Multiply $- 3$ by $2$ .

$x = \frac{1 - 6}{2}$

Subtract $6$ from $1$ .

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

Move the negative in front of the fraction.

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

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = - 2.5$

Mixed Number Form:

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