Solve 1/16=128{left(1/2right)}^n-1

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

Solve for n

$$n=12$$

Steps Using Definition of Logarithm

Divide both sides by 128.

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

Express \frac{\frac{1}{16}}{128} as a single fraction.

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

Multiply 16 and 128 to get 2048.

$$\frac{1}{2048}=\left(\frac{1}{2}\right)^{n-1}$$

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

$$\left(\frac{1}{2}\right)^{n-1}=\frac{1}{2048}$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{1}{2}\right)^{n-1})=\log(\frac{1}{2048})$$

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

$$\left(n-1\right)\log(\frac{1}{2})=\log(\frac{1}{2048})$$

Divide both sides by \log(\frac{1}{2}).

$$n-1=\frac{\log(\frac{1}{2048})}{\log(\frac{1}{2})}$$

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

$$n-1=\log_{\frac{1}{2}}\left(\frac{1}{2048}\right)$$

Add 1 to both sides of the equation.

$$n=11-\left(-1\right)$$