Solve 7^4x+11=1/7

$$7 ^ { 4 x + 11 } = \frac { 1 } { 7 }$$

Solve for x

$$x=-3$$

Steps Using Definition of Logarithm

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

$$7^{4x+11}=\frac{1}{7}$$

Take the logarithm of both sides of the equation.

$$\log(7^{4x+11})=\log(\frac{1}{7})$$

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

$$\left(4x+11\right)\log(7)=\log(\frac{1}{7})$$

Divide both sides by \log(7).

$$4x+11=\frac{\log(\frac{1}{7})}{\log(7)}$$

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

$$4x+11=\log_{7}\left(\frac{1}{7}\right)$$

Subtract 11 from both sides of the equation.

$$4x=-1-11$$

Divide both sides by 4.

$$x=-\frac{12}{4}$$

Solve for x (complex solution)

$$x=\frac{i\pi n_{1}}{2\ln(7)}-3$$

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