Solve (7^2)^9=(7^3)^x

$$( 7 ^ { 2 } ) ^ { 9 } = ( 7 ^ { 3 } ) ^ { x }$$

Solve for x

$$x=6$$

Steps Using Definition of Logarithm

To raise a power to another power, multiply the exponents. Multiply 2 and 9 to get 18.

$$7^{18}=\left(7^{3}\right)^{x}$$

Calculate 7 to the power of 18 and get 1628413597910449.

$$1628413597910449=\left(7^{3}\right)^{x}$$

Calculate 7 to the power of 3 and get 343.

$$1628413597910449=343^{x}$$

Swap sides so that all variable terms are on the left hand side.

$$343^{x}=1628413597910449$$

Take the logarithm of both sides of the equation.

$$\log(343^{x})=\log(1628413597910449)$$

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

$$x\log(343)=\log(1628413597910449)$$

Divide both sides by \log(343).

$$x=\frac{\log(1628413597910449)}{\log(343)}$$

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

$$x=\log_{343}\left(1628413597910449\right)$$

Solve for x (complex solution)

$$x=\frac{2\pi n_{1}i}{3\ln(7)}+6$$

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