Solve (7^2)^n=49.7^8

$$( 7 ^ { 2 } ) ^ { n } = 49.7 ^ { 8 }$$

Solve for n

$$n=4\log_{7}\left(49.7\right)\approx 8.029157842$$

Steps Using Definition of Logarithm

Calculate 7 to the power of 2 and get 49.

$$49^{n}=49.7^{8}$$

Calculate 49.7 to the power of 8 and get 37226406026790.94258561.

$$49^{n}=37226406026790.94258561$$

Take the logarithm of both sides of the equation.

$$\log(49^{n})=\log(37226406026790.94258561)$$

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

$$n\log(49)=\log(37226406026790.94258561)$$

Divide both sides by \log(49).

$$n=\frac{\log(37226406026790.94258561)}{\log(49)}$$

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

$$n=\log_{49}\left(37226406026790.94258561\right)$$