Solve (1/16)^3x-2=1/128

$$( \frac { 1 } { 16 } ) ^ { 3 x - 2 } = \frac { 1 } { 128 }$$

Solve for x

$$x = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

Steps Using Definition of Logarithm

Use the rules of exponents and logarithms to solve the equation.

$$\left(\frac{1}{16}\right)^{3x-2}=\frac{1}{128}$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{1}{16}\right)^{3x-2})=\log(\frac{1}{128})$$

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

$$\left(3x-2\right)\log(\frac{1}{16})=\log(\frac{1}{128})$$

Divide both sides by \log(\frac{1}{16}).

$$3x-2=\frac{\log(\frac{1}{128})}{\log(\frac{1}{16})}$$

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

$$3x-2=\log_{\frac{1}{16}}\left(\frac{1}{128}\right)$$

Add 2 to both sides of the equation.

$$3x=\frac{7}{4}-\left(-2\right)$$

Divide both sides by 3.

$$x=\frac{\frac{15}{4}}{3}$$

Solve for x (complex solution)

$$x=-\frac{i\pi n_{1}}{6\ln(2)}+\frac{5}{4}$$

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