Solve (7^2)^n=497^8

$$( 7 ^ { 2 } ) ^ { n } = 497 ^ { 8 }$$

Solve for n

$$n=4\log_{7}\left(497\right)\approx 12.762336491$$

Steps Using Definition of Logarithm

Calculate 7 to the power of 2 and get 49.

$$49^{n}=497^{8}$$

Calculate 497 to the power of 8 and get 3722640602679094258561.

$$49^{n}=3722640602679094258561$$

Take the logarithm of both sides of the equation.

$$\log(49^{n})=\log(3722640602679094258561)$$

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

$$n\log(49)=\log(3722640602679094258561)$$

Divide both sides by \log(49).

$$n=\frac{\log(3722640602679094258561)}{\log(49)}$$

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

$$n=\log_{49}\left(3722640602679094258561\right)$$