Solve for x 25^2=125^(x+1)

$25^{2} = 125^{x + 1}$

Step 1

Rewrite the equation as $125^{x + 1} = 25^{2}$ .

$125^{x + 1} = 25^{2}$

Step 2

Create equivalent expressions in the equation that all have equal bases.

$5^{3 (x + 1)} = 5^{2 \cdot 2}$

Step 3

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

$3 (x + 1) = 2 \cdot 2$

Step 4

Solve for $x$ .

Multiply $2$ by $2$ .

$3 (x + 1) = 4$

Divide each term in $3 (x + 1) = 4$ by $3$ and simplify.

Divide each term in $3 (x + 1) = 4$ by $3$ .

$\frac{3 (x + 1)}{3} = \frac{4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 (x + 1)}{3} = \frac{4}{3}$

Divide $x + 1$ by $1$ .

$x + 1 = \frac{4}{3}$

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

Subtract $1$ from both sides of the equation.

$x = \frac{4}{3} - 1$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$x = \frac{4}{3} - 1 \cdot \frac{3}{3}$

Combine $- 1$ and $\frac{3}{3}$ .

$x = \frac{4}{3} + \frac{- 1 \cdot 3}{3}$

Combine the numerators over the common denominator.

$x = \frac{4 - 1 \cdot 3}{3}$

Simplify the numerator.

Multiply $- 1$ by $3$ .

$x = \frac{4 - 3}{3}$

Subtract $3$ from $4$ .

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

Step 5

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 0.\overline{3}$