Solve (2^5)=2^b(1)

$$( 2 ^ { 5 } ) = 2 ^ { b } ( 1 )$$

Solve for b

$$b=5$$

Steps Using Definition of Logarithm

Calculate 2 to the power of 5 and get 32.

$$32=2^{b}\times 1$$

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

$$2^{b}\times 1=32$$

Divide both sides by 1.

$$2^{b}=32$$

Take the logarithm of both sides of the equation.

$$\log(2^{b})=\log(32)$$

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

$$b\log(2)=\log(32)$$

Divide both sides by \log(2).

$$b=\frac{\log(32)}{\log(2)}$$

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

$$b=\log_{2}\left(32\right)$$

Solve for b (complex solution)

$$b=\frac{2\pi n_{1}i}{\ln(2)}+5$$

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