Solve (7^2)^-x=7

$$( 7 ^ { 2 } ) ^ { - x } = 7$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

Steps Using Definition of Logarithm

Calculate 7 to the power of 2 and get 49.

$$49^{-x}=7$$

Take the logarithm of both sides of the equation.

$$\log(49^{-x})=\log(7)$$

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

$$-x\log(49)=\log(7)$$

Divide both sides by \log(49).

$$-x=\frac{\log(7)}{\log(49)}$$

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

$$-x=\log_{49}\left(7\right)$$

Divide both sides by -1.

$$x=\frac{\frac{1}{2}}{-1}$$

Solve for x (complex solution)

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

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