Solve 3^3x+1-4=2

$$3 ^ { 3 x + 1 } - 4 = 2$$

Solve for x

$$x=\frac{\log_{3}\left(2\right)}{3}\approx 0.210309918$$

Steps Using Definition of Logarithm

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

$$3^{3x+1}-4=2$$

Add 4 to both sides of the equation.

$$3^{3x+1}=6$$

Take the logarithm of both sides of the equation.

$$\log(3^{3x+1})=\log(6)$$

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

$$\left(3x+1\right)\log(3)=\log(6)$$

Divide both sides by \log(3).

$$3x+1=\frac{\log(6)}{\log(3)}$$

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

$$3x+1=\log_{3}\left(6\right)$$

Subtract 1 from both sides of the equation.

$$3x=\log_{3}\left(6\right)-1$$

Divide both sides by 3.

$$x=\frac{\log_{3}\left(6\right)-1}{3}$$

Solve for x (complex solution)

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

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