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

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

Step 1

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

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

Step 2

Solve for $x$ .

Rewrite the equation as $4 + \log_{2} (x) = 3^{2}$ .

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

Raise $3$ to the power of $2$ .

$4 + \log_{2} (x) = 9$

Move all terms not containing $\log_{2} (x)$ to the right side of the equation.

Subtract $4$ from both sides of the equation.

$\log_{2} (x) = 9 - 4$

Subtract $4$ from $9$ .

$\log_{2} (x) = 5$

Rewrite $\log_{2} (x) = 5$ 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 $b^{y} = x$ .

$2^{5} = x$

Solve for $x$ .

Rewrite the equation as $x = 2^{5}$ .

$x = 2^{5}$

Raise $2$ to the power of $5$ .

$x = 32$

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

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