Solve for x log base 4 of 3x+4=2

$\log_{4} (3 x + 4) = 2$

Step 1

Rewrite $\log_{4} (3 x + 4) = 2$ 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 $b^{y} = x$ .

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

Step 2

Solve for $x$ .

Rewrite the equation as $3 x + 4 = 4^{2}$ .

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

Raise $4$ to the power of $2$ .

$3 x + 4 = 16$

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

Subtract $4$ from both sides of the equation.

$3 x = 16 - 4$

Subtract $4$ from $16$ .

$3 x = 12$

Divide each term in $3 x = 12$ by $3$ and simplify.

Divide each term in $3 x = 12$ by $3$ .

$\frac{3 x}{3} = \frac{12}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{12}{3}$

Divide $x$ by $1$ .

$x = \frac{12}{3}$

Simplify the right side.

Divide $12$ by $3$ .

$x = 4$

Solve for x log base 4 of 3x+4=2

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