Solve (1/2)^2=(1/4)^3r+1

$$( \frac { 1 } { 2 } ) ^ { 2 } = ( \frac { 1 } { 4 } ) ^ { 3 r + 1 }$$

Solve for r

$$r=0$$

Steps Using Definition of Logarithm

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$\frac{1}{4}=\left(\frac{1}{4}\right)^{3r+1}$$

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

$$\left(\frac{1}{4}\right)^{3r+1}=\frac{1}{4}$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{1}{4}\right)^{3r+1})=\log(\frac{1}{4})$$

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

$$\left(3r+1\right)\log(\frac{1}{4})=\log(\frac{1}{4})$$

Divide both sides by \log(\frac{1}{4}).

$$3r+1=\frac{\log(\frac{1}{4})}{\log(\frac{1}{4})}$$

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

$$3r+1=\log_{\frac{1}{4}}\left(\frac{1}{4}\right)$$

Subtract 1 from both sides of the equation.

$$3r=1-1$$

Divide both sides by 3.

$$r=\frac{0}{3}$$