Solve for x 1/128=2^(5x+3)

$\frac{1}{128} = 2^{5 x + 3}$

Step 1

Rewrite the equation as $2^{5 x + 3} = \frac{1}{128}$ .

$2^{5 x + 3} = \frac{1}{128}$

Step 2

Raise $128$ to the power of $1$ .

$2^{5 x + 3} = \frac{1}{128^{1}}$

Step 3

Move $128^{1}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$2^{5 x + 3} = 128^{- 1}$

Step 4

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

$2^{5 x + 3} = 2^{- 7}$

Step 5

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

$5 x + 3 = - 7$

Step 6

Solve for $x$ .

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

Subtract $3$ from both sides of the equation.

$5 x = - 7 - 3$

Subtract $3$ from $- 7$ .

$5 x = - 10$

Divide each term in $5 x = - 10$ by $5$ and simplify.

Divide each term in $5 x = - 10$ by $5$ .

$\frac{5 x}{5} = \frac{- 10}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{- 10}{5}$

Divide $x$ by $1$ .

$x = \frac{- 10}{5}$

Simplify the right side.

Divide $- 10$ by $5$ .

$x = - 2$