Solve for x 4^(4x-5)=8^(3x-4)

$4^{4 x - 5} = 8^{3 x - 4}$

Step 1

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

$2^{2 (4 x - 5)} = 2^{3 (3 x - 4)}$

Step 2

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

$2 (4 x - 5) = 3 (3 x - 4)$

Step 3

Solve for $x$ .

Simplify $2 (4 x - 5)$ .

Rewrite.

$0 + 0 + 2 (4 x - 5) = 3 (3 x - 4)$

Simplify by adding zeros.

$2 (4 x - 5) = 3 (3 x - 4)$

Apply the distributive property.

$2 (4 x) + 2 \cdot - 5 = 3 (3 x - 4)$

Multiply.

Multiply $4$ by $2$ .

$8 x + 2 \cdot - 5 = 3 (3 x - 4)$

Multiply $2$ by $- 5$ .

$8 x - 10 = 3 (3 x - 4)$

Simplify $3 (3 x - 4)$ .

Apply the distributive property.

$8 x - 10 = 3 (3 x) + 3 \cdot - 4$

Multiply.

Multiply $3$ by $3$ .

$8 x - 10 = 9 x + 3 \cdot - 4$

Multiply $3$ by $- 4$ .

$8 x - 10 = 9 x - 12$

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

Subtract $9 x$ from both sides of the equation.

$8 x - 10 - 9 x = - 12$

Subtract $9 x$ from $8 x$ .

$- x - 10 = - 12$

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

Add $10$ to both sides of the equation.

$- x = - 12 + 10$

Add $- 12$ and $10$ .

$- x = - 2$

Divide each term in $- x = - 2$ by $- 1$ and simplify.

Divide each term in $- x = - 2$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 2}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 2}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 2}{- 1}$

Simplify the right side.

Divide $- 2$ by $- 1$ .

$x = 2$

Solve for x 4^(4x-5)=8^(3x-4)

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