Solve for x log of 3x+1=2

$\log (3 x + 1) = 2$

Rewrite $\log (3 x + 1) = 2$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$10^{2} = 3 x + 1$

Solve for $x$ .

Rewrite the equation as $3 x + 1 = 10^{2}$ .

$3 x + 1 = 10^{2}$

Raise $10$ to the power of $2$ .

$3 x + 1 = 100$

Move all terms not containing $x$ to the right side of the equation.

Subtract $1$ from both sides of the equation.

$3 x = 100 - 1$

Subtract $1$ from $100$ .

$3 x = 99$

Divide each term in $3 x = 99$ by $3$ and simplify.

Divide each term in $3 x = 99$ by $3$ .

$\frac{3 x}{3} = \frac{99}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{99}{3}$

Divide $x$ by $1$ .

$x = \frac{99}{3}$

Simplify the right side.

Divide $99$ by $3$ .

$x = 33$

Solve for x log of 3x+1=2

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