Solve for c 1/5=5^(2c+3)

$\frac{1}{5} = 5^{2 c + 3}$

Rewrite the equation as $5^{2 c + 3} = \frac{1}{5}$ .

$5^{2 c + 3} = \frac{1}{5}$

Raise $5$ to the power of $1$ .

$5^{2 c + 3} = \frac{1}{5^{1}}$

Move $5^{1}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$5^{2 c + 3} = 5^{- 1}$

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

$2 c + 3 = - 1$

Solve for $c$ .

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

Subtract $3$ from both sides of the equation.

$2 c = - 1 - 3$

Subtract $3$ from $- 1$ .

$2 c = - 4$

Divide each term in $2 c = - 4$ by $2$ and simplify.

Divide each term in $2 c = - 4$ by $2$ .

$\frac{2 c}{2} = \frac{- 4}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 c}{2} = \frac{- 4}{2}$

Divide $c$ by $1$ .

$c = \frac{- 4}{2}$

Simplify the right side.

Divide $- 4$ by $2$ .

$c = - 2$