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

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

Step 1

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

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

Step 2

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

$5^{2 (2 h + 1)} = 5^{3}$

Step 3

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

$2 (2 h + 1) = 3$

Step 4

Solve for $h$ .

Divide each term in $2 (2 h + 1) = 3$ by $2$ and simplify.

Divide each term in $2 (2 h + 1) = 3$ by $2$ .

$\frac{2 (2 h + 1)}{2} = \frac{3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 (2 h + 1)}{2} = \frac{3}{2}$

Divide $2 h + 1$ by $1$ .

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

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

Subtract $1$ from both sides of the equation.

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

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Combine $- 1$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $- 1$ by $2$ .

$2 h = \frac{3 - 2}{2}$

Subtract $2$ from $3$ .

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

Divide each term in $2 h = \frac{1}{2}$ by $2$ and simplify.

Divide each term in $2 h = \frac{1}{2}$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $h$ by $1$ .

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

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$h = \frac{1}{2} \cdot \frac{1}{2}$

Multiply $\frac{1}{2} \cdot \frac{1}{2}$ .

Multiply $\frac{1}{2}$ by $\frac{1}{2}$ .

$h = \frac{1}{2 \cdot 2}$

Multiply $2$ by $2$ .

$h = \frac{1}{4}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$h = \frac{1}{4}$

Decimal Form:

$h = 0.25$

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

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