Solve 4^4x+4=9

$$4 ^ { 4 x + 4 } = 9$$

Solve for x

$$x=\frac{\log_{2}\left(3\right)-4}{4}\approx -0.603759375$$

Steps Using Definition of Logarithm

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

$$4^{4x+4}=9$$

Take the logarithm of both sides of the equation.

$$\log(4^{4x+4})=\log(9)$$

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

$$\left(4x+4\right)\log(4)=\log(9)$$

Divide both sides by \log(4).

$$4x+4=\frac{\log(9)}{\log(4)}$$

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

$$4x+4=\log_{4}\left(9\right)$$

Subtract 4 from both sides of the equation.

$$4x=\log_{2}\left(3\right)-4$$

Divide both sides by 4.

$$x=\frac{\log_{2}\left(3\right)-4}{4}$$