Solve 125^9x-2=150

$${ 125 }^{ 9x-2 } =150$$

Solve for x

$$x=\frac{\log_{5}\left(6\right)+8}{27}\approx 0.337528991$$

Steps Using Definition of Logarithm

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

$$125^{9x-2}=150$$

Take the logarithm of both sides of the equation.

$$\log(125^{9x-2})=\log(150)$$

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

$$\left(9x-2\right)\log(125)=\log(150)$$

Divide both sides by \log(125).

$$9x-2=\frac{\log(150)}{\log(125)}$$

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

$$9x-2=\log_{125}\left(150\right)$$

Add 2 to both sides of the equation.

$$9x=\frac{\log_{5}\left(150\right)}{3}-\left(-2\right)$$

Divide both sides by 9.

$$x=\frac{\frac{\log_{5}\left(150\right)}{3}+2}{9}$$

Solve for x (complex solution)

$$x=\frac{2\pi n_{1}i}{27\ln(5)}+\frac{\log_{5}\left(6\right)}{27}+\frac{8}{27}$$

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