Solve 1/32=2^1-m

$$\frac { 1 } { 32 } = 2 ^ { 1 - m }$$

Solve for m

$$m=6$$

Steps Using Definition of Logarithm

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

$$2^{1-m}=\frac{1}{32}$$

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

$$2^{-m+1}=\frac{1}{32}$$

Take the logarithm of both sides of the equation.

$$\log(2^{-m+1})=\log(\frac{1}{32})$$

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

$$\left(-m+1\right)\log(2)=\log(\frac{1}{32})$$

Divide both sides by \log(2).

$$-m+1=\frac{\log(\frac{1}{32})}{\log(2)}$$

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

$$-m+1=\log_{2}\left(\frac{1}{32}\right)$$

Subtract 1 from both sides of the equation.

$$-m=-5-1$$

Divide both sides by -1.

$$m=-\frac{6}{-1}$$