Solve 9*3^x=7*3^x+18

$$9 \cdot 3 ^ { x } = 7 \cdot 3 ^ { x } + 18$$

Solve for x

$$x=2$$

Steps Using Definition of Logarithm

Subtract 7\times 3^{x} from both sides.

$$9\times 3^{x}-7\times 3^{x}=18$$

Combine 9\times 3^{x} and -7\times 3^{x} to get 2\times 3^{x}.

$$2\times 3^{x}=18$$

Divide both sides by 2.

$$3^{x}=\frac{18}{2}$$

Divide 18 by 2 to get 9.

$$3^{x}=9$$

Take the logarithm of both sides of the equation.

$$\log(3^{x})=\log(9)$$

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

$$x\log(3)=\log(9)$$

Divide both sides by \log(3).

$$x=\frac{\log(9)}{\log(3)}$$

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

$$x=\log_{3}\left(9\right)$$

Solve for x (complex solution)

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

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