Solve for x 3^(2x)=g^(3x-4)

$3^{2 x} = g^{3 x - 4}$

Step 1

Take the log of both sides of the equation.

$\ln (3^{2 x}) = \ln (g^{3 x - 4})$

Step 2

Expand $\ln (3^{2 x})$ by moving $2 x$ outside the logarithm.

$2 x \ln (3) = \ln (g^{3 x - 4})$

Step 3

Expand $\ln (g^{3 x - 4})$ by moving $3 x - 4$ outside the logarithm.

$2 x \ln (3) = (3 x - 4) \ln (g)$

Step 4

Solve the equation for $x$ .

Apply the distributive property.

$2 x \ln (3) = 3 x \ln (g) - 4 \ln (g)$

Subtract $3 x \ln (g)$ from both sides of the equation.

$2 x \ln (3) - 3 x \ln (g) = - 4 \ln (g)$

Factor $x$ out of $2 x \ln (3) - 3 x \ln (g)$ .

Factor $x$ out of $2 x \ln (3)$ .

$x (2 \ln (3)) - 3 x \ln (g) = - 4 \ln (g)$

Factor $x$ out of $- 3 x \ln (g)$ .

$x (2 \ln (3)) + x (- 3 \ln (g)) = - 4 \ln (g)$

Factor $x$ out of $x (2 \ln (3)) + x (- 3 \ln (g))$ .

$x (2 \ln (3) - 3 \ln (g)) = - 4 \ln (g)$

Divide each term in $x (2 \ln (3) - 3 \ln (g)) = - 4 \ln (g)$ by $2 \ln (3) - 3 \ln (g)$ and simplify.

Divide each term in $x (2 \ln (3) - 3 \ln (g)) = - 4 \ln (g)$ by $2 \ln (3) - 3 \ln (g)$ .

$\frac{x (2 \ln (3) - 3 \ln (g))}{2 \ln (3) - 3 \ln (g)} = \frac{- 4 \ln (g)}{2 \ln (3) - 3 \ln (g)}$

Simplify the left side.

Cancel the common factor of $2 \ln (3) - 3 \ln (g)$ .

Cancel the common factor.

$\frac{x (2 \ln (3) - 3 \ln (g))}{2 \ln (3) - 3 \ln (g)} = \frac{- 4 \ln (g)}{2 \ln (3) - 3 \ln (g)}$

Divide $x$ by $1$ .

$x = \frac{- 4 \ln (g)}{2 \ln (3) - 3 \ln (g)}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{4 \ln (g)}{2 \ln (3) - 3 \ln (g)}$