Solve 9^-3x+2=48

$${ 9 }^{ -3x+2 } =48$$

Solve for x

$$x=-\frac{2\log_{3}\left(2\right)}{3}+\frac{1}{2}\approx 0.079380164$$

Steps Using Definition of Logarithm

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

$$9^{-3x+2}=48$$

Take the logarithm of both sides of the equation.

$$\log(9^{-3x+2})=\log(48)$$

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

$$\left(-3x+2\right)\log(9)=\log(48)$$

Divide both sides by \log(9).

$$-3x+2=\frac{\log(48)}{\log(9)}$$

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

$$-3x+2=\log_{9}\left(48\right)$$

Subtract 2 from both sides of the equation.

$$-3x=\frac{\log_{3}\left(48\right)}{2}-2$$

Divide both sides by -3.

$$x=\frac{\frac{\log_{3}\left(48\right)}{2}-2}{-3}$$

Solve for x (complex solution)

$$x=-\frac{\pi n_{1}i}{3\ln(3)}-\frac{2\log_{3}\left(2\right)}{3}+\frac{1}{2}$$

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