Solve 1/128=2^5x+3

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

Solve for x

$$x=-2$$

Steps Using Definition of Logarithm

Swap sides so that all variable terms are on the left hand side.

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

Take the logarithm of both sides of the equation.

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

The logarithm of a number raised to a power is the power times the logarithm of the number.

$$\left(5x+3\right)\log(2)=\log(\frac{1}{128})$$

Divide both sides by \log(2).

$$5x+3=\frac{\log(\frac{1}{128})}{\log(2)}$$

By the change-of-base formula \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).

$$5x+3=\log_{2}\left(\frac{1}{128}\right)$$

Subtract 3 from both sides of the equation.

$$5x=-7-3$$

Divide both sides by 5.

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

Solve for x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{5\ln(2)}-2$$

$$n_{1}\in \mathrm{Z}$$