Solve 9^x+4=(36)^3*(8/3)^-2

$$9 ^ { x + 4 } = ( 36 ) ^ { 3 } \times ( \frac { 8 } { 3 } ) ^ { - 2 }$$

Solve for x

$$x=0$$

Steps Using Definition of Logarithm

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

$$9^{x+4}=6561$$

Take the logarithm of both sides of the equation.

$$\log(9^{x+4})=\log(6561)$$

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

$$\left(x+4\right)\log(9)=\log(6561)$$

Divide both sides by \log(9).

$$x+4=\frac{\log(6561)}{\log(9)}$$

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

$$x+4=\log_{9}\left(6561\right)$$

Subtract 4 from both sides of the equation.

$$x=4-4$$

Solve for x (complex solution)

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

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