Solve for k 2* log base 4 of k=4

$2 \cdot \log_{4} (k) = 4$

Divide each term in $(2 \cdot \log_{4} (k)) = 4$ by $2$ and simplify.

Divide each term in $(2 \cdot \log_{4} (k)) = 4$ by $2$ .

$\frac{(2 \cdot \log_{4} (k))}{2} = \frac{4}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 \cdot \log_{4} (k)}{2} = \frac{4}{2}$

Divide $\log_{4} (k)$ by $1$ .

$\log_{4} (k) = \frac{4}{2}$

Simplify the right side.

Divide $4$ by $2$ .

$\log_{4} (k) = 2$

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

$4^{2} = k$

Solve for $k$ .

Rewrite the equation as $k = 4^{2}$ .

$k = 4^{2}$

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

$k = 16$

Solve for k 2* log base 4 of k=4

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