Solve 3^-3x+4=1/9?

$$3 ^ { - 3 x + 4 } = \frac { 1 } { 9 } ?$$

Solve for x

$$x=2$$

Steps Using Definition of Logarithm

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

$$3^{-3x+4}=\frac{1}{9}$$

Take the logarithm of both sides of the equation.

$$\log(3^{-3x+4})=\log(\frac{1}{9})$$

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

$$\left(-3x+4\right)\log(3)=\log(\frac{1}{9})$$

Divide both sides by \log(3).

$$-3x+4=\frac{\log(\frac{1}{9})}{\log(3)}$$

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

$$-3x+4=\log_{3}\left(\frac{1}{9}\right)$$

Subtract 4 from both sides of the equation.

$$-3x=-2-4$$

Divide both sides by -3.

$$x=-\frac{6}{-3}$$

Solve for x (complex solution)

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

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