Solve 2^y+3^x=18

$$2 ^ { y + 3 ^ { x } } = 18$$

Solve for y

$$y=-3^{x}+2\log_{2}\left(3\right)+1$$

Steps Using Definition of Logarithm

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

$$2^{y+3^{x}}=18$$

Take the logarithm of both sides of the equation.

$$\log(2^{y+3^{x}})=\log(18)$$

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

$$\left(y+3^{x}\right)\log(2)=\log(18)$$

Divide both sides by \log(2).

$$y+3^{x}=\frac{\log(18)}{\log(2)}$$

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

$$y+3^{x}=\log_{2}\left(18\right)$$

Subtract 3^{x} from both sides of the equation.

$$y=\log_{2}\left(18\right)-3^{x}$$

Solve for x

$$x=\log_{3}\left(-y+2\log_{2}\left(3\right)+1\right)$$

$$y<2\log_{2}\left(3\right)+1$$