Solve 7=9^4x+1

$$7 = 9 ^ { 4 x + 1 }$$

Solve for x

$$x=\frac{\log_{3}\left(7\right)-2}{8}\approx -0.028594531$$

Steps Using Definition of Logarithm

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

$$9^{4x+1}=7$$

Take the logarithm of both sides of the equation.

$$\log(9^{4x+1})=\log(7)$$

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

$$\left(4x+1\right)\log(9)=\log(7)$$

Divide both sides by \log(9).

$$4x+1=\frac{\log(7)}{\log(9)}$$

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

$$4x+1=\log_{9}\left(7\right)$$

Subtract 1 from both sides of the equation.

$$4x=\frac{\log_{3}\left(7\right)}{2}-1$$

Divide both sides by 4.

$$x=\frac{\frac{\log_{3}\left(7\right)}{2}-1}{4}$$

Solve for x (complex solution)

$$x=\frac{\pi n_{1}i}{4\ln(3)}+\frac{\log_{3}\left(7\right)}{8}-\frac{1}{4}$$

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