Solve (2^-1)^n=2^(-6)

$$( 2 ^ { - 1 } ) ^ { n } = 2 ^ { ( - 6 ) }$$

Solve for n

$$n=6$$

Steps Using Definition of Logarithm

Calculate 2 to the power of -1 and get \frac{1}{2}.

$$\left(\frac{1}{2}\right)^{n}=2^{-6}$$

Calculate 2 to the power of -6 and get \frac{1}{64}.

$$\left(\frac{1}{2}\right)^{n}=\frac{1}{64}$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{1}{2}\right)^{n})=\log(\frac{1}{64})$$

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

$$n\log(\frac{1}{2})=\log(\frac{1}{64})$$

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

$$n=\frac{\log(\frac{1}{64})}{\log(\frac{1}{2})}$$

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

$$n=\log_{\frac{1}{2}}\left(\frac{1}{64}\right)$$