Solve 63^2x-2=81

$$63 ^ { 2 x - 2 } = 81$$

Solve for x

$$x=\log_{63}\left(567\right)\approx 1.53032902$$

Steps Using Definition of Logarithm

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

$$63^{2x-2}=81$$

Take the logarithm of both sides of the equation.

$$\log(63^{2x-2})=\log(81)$$

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

$$\left(2x-2\right)\log(63)=\log(81)$$

Divide both sides by \log(63).

$$2x-2=\frac{\log(81)}{\log(63)}$$

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

$$2x-2=\log_{63}\left(81\right)$$

Add 2 to both sides of the equation.

$$2x=4\log_{63}\left(3\right)-\left(-2\right)$$

Divide both sides by 2.

$$x=\frac{4\log_{63}\left(3\right)+2}{2}$$