Solve for c 5^(c-2)=8^(2c)

$5^{c - 2} = 8^{2 c}$

Take the log of both sides of the equation.

$\ln (5^{c - 2}) = \ln (8^{2 c})$

Expand $\ln (5^{c - 2})$ by moving $c - 2$ outside the logarithm.

$(c - 2) \ln (5) = \ln (8^{2 c})$

Expand $\ln (8^{2 c})$ by moving $2 c$ outside the logarithm.

$(c - 2) \ln (5) = 2 c \ln (8)$

Solve the equation for $c$ .

Simplify $(c - 2) \ln (5)$ .

Start simplifying.

$0 + 0 + (c - 2) \ln (5) = 2 c \ln (8)$

Simplify by adding zeros.

$(c - 2) \ln (5) = 2 c \ln (8)$

Apply the distributive property.

$c \ln (5) - 2 \ln (5) = 2 c \ln (8)$

Subtract $2 c \ln (8)$ from both sides of the equation.

$c \ln (5) - 2 \ln (5) - 2 c \ln (8) = 0$

Add $2 \ln (5)$ to both sides of the equation.

$c \ln (5) - 2 c \ln (8) = 2 \ln (5)$

Factor $c$ out of $c \ln (5) - 2 c \ln (8)$ .

Factor $c$ out of $c \ln (5)$ .

$c (\ln (5)) - 2 c \ln (8) = 2 \ln (5)$

Factor $c$ out of $- 2 c \ln (8)$ .

$c (\ln (5)) + c (- 2 \ln (8)) = 2 \ln (5)$

Factor $c$ out of $c (\ln (5)) + c (- 2 \ln (8))$ .

$c (\ln (5) - 2 \ln (8)) = 2 \ln (5)$

Divide each term in $c (\ln (5) - 2 \ln (8)) = 2 \ln (5)$ by $\ln (5) - 2 \ln (8)$ and simplify.

Divide each term in $c (\ln (5) - 2 \ln (8)) = 2 \ln (5)$ by $\ln (5) - 2 \ln (8)$ .

$\frac{c (\ln (5) - 2 \ln (8))}{\ln (5) - 2 \ln (8)} = \frac{2 \ln (5)}{\ln (5) - 2 \ln (8)}$

Simplify the left side.

Cancel the common factor of $\ln (5) - 2 \ln (8)$ .

Cancel the common factor.

$\frac{c (\ln (5) - 2 \ln (8))}{\ln (5) - 2 \ln (8)} = \frac{2 \ln (5)}{\ln (5) - 2 \ln (8)}$

Divide $c$ by $1$ .

$c = \frac{2 \ln (5)}{\ln (5) - 2 \ln (8)}$

The result can be shown in multiple forms.

Exact Form:

$c = \frac{2 \ln (5)}{\ln (5) - 2 \ln (8)}$

Decimal Form:

$c = - 1.26257895 \dots$