Solve for x 243^(4x+5)=(1/243)^(-3x+13)

$243^{4 x + 5} = {(\frac{1}{243})}^{- 3 x + 13}$

Apply the product rule to $\frac{1}{243}$ .

$243^{4 x + 5} = \frac{1^{- 3 x + 13}}{243^{- 3 x + 13}}$

One to any power is one.

$243^{4 x + 5} = \frac{1}{243^{- 3 x + 13}}$

Move $243^{- 3 x + 13}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$243^{4 x + 5} = 243^{- (- 3 x + 13)}$

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

$4 x + 5 = - (- 3 x + 13)$

Solve for $x$ .

Simplify $- (- 3 x + 13)$ .

Rewrite.

$4 x + 5 = 0 + 0 - (- 3 x + 13)$

Simplify by adding zeros.

$4 x + 5 = - (- 3 x + 13)$

Apply the distributive property.

$4 x + 5 = - (- 3 x) - 1 \cdot 13$

Multiply.

Multiply $- 3$ by $- 1$ .

$4 x + 5 = 3 x - 1 \cdot 13$

Multiply $- 1$ by $13$ .

$4 x + 5 = 3 x - 13$

Move all terms containing $x$ to the left side of the equation.

Subtract $3 x$ from both sides of the equation.

$4 x + 5 - 3 x = - 13$

Subtract $3 x$ from $4 x$ .

$x + 5 = - 13$

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

Subtract $5$ from both sides of the equation.

$x = - 13 - 5$

Subtract $5$ from $- 13$ .

$x = - 18$