Solve for n log base 4 of n=1/4 log base 4 of 81+1/2 log base 4 of 25

$\log_{4} (n) = \frac{1}{4} \log_{4} (81) + \frac{1}{2} \log_{4} (25)$

Simplify the right side.

Simplify each term.

Combine $\frac{1}{4}$ and $\log_{4} (81)$ .

$\log_{4} (n) = \frac{\log_{4} (81)}{4} + \frac{1}{2} \log_{4} (25)$

Combine $\frac{1}{2}$ and $\log_{4} (25)$ .

$\log_{4} (n) = \frac{\log_{4} (81)}{4} + \frac{\log_{4} (25)}{2}$

Multiply each term in $\log_{4} (n) = \frac{\log_{4} (81)}{4} + \frac{\log_{4} (25)}{2}$ by $4$ to eliminate the fractions.

Multiply each term in $\log_{4} (n) = \frac{\log_{4} (81)}{4} + \frac{\log_{4} (25)}{2}$ by $4$ .

$\log_{4} (n) \cdot 4 = \frac{\log_{4} (81)}{4} \cdot 4 + \frac{\log_{4} (25)}{2} \cdot 4$

Simplify the left side.

Move $4$ to the left of $\log_{4} (n)$ .

$4 \log_{4} (n) = \frac{\log_{4} (81)}{4} \cdot 4 + \frac{\log_{4} (25)}{2} \cdot 4$

Simplify the right side.

Simplify each term.

Cancel the common factor of $4$ .

Cancel the common factor.

$4 \log_{4} (n) = \frac{\log_{4} (81)}{4} \cdot 4 + \frac{\log_{4} (25)}{2} \cdot 4$

Rewrite the expression.

$4 \log_{4} (n) = \log_{4} (81) + \frac{\log_{4} (25)}{2} \cdot 4$

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

$4 \log_{4} (n) = \log_{4} (81) + \frac{\log_{4} (25)}{2} \cdot (2 (2))$

Cancel the common factor.

$4 \log_{4} (n) = \log_{4} (81) + \frac{\log_{4} (25)}{2} \cdot (2 \cdot 2)$

Rewrite the expression.

$4 \log_{4} (n) = \log_{4} (81) + \log_{4} (25) \cdot 2$

Move $2$ to the left of $\log_{4} (25)$ .

$4 \log_{4} (n) = \log_{4} (81) + 2 \log_{4} (25)$

Simplify the left side.

Simplify $4 \log_{4} (n)$ by moving $4$ inside the logarithm.

$\log_{4} (n^{4}) = \log_{4} (81) + 2 \log_{4} (25)$

Simplify the right side.

Simplify $\log_{4} (81) + 2 \log_{4} (25)$ .

Simplify each term.

Simplify $2 \log_{4} (25)$ by moving $2$ inside the logarithm.

$\log_{4} (n^{4}) = \log_{4} (81) + \log_{4} (25^{2})$

Raise $25$ to the power of $2$ .

$\log_{4} (n^{4}) = \log_{4} (81) + \log_{4} (625)$

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log_{4} (n^{4}) = \log_{4} (81 \cdot 625)$

Multiply $81$ by $625$ .

$\log_{4} (n^{4}) = \log_{4} (50625)$

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$n^{4} = 50625$

Solve for $n$ .

Take the 4th root of both sides of the equation to eliminate the exponent on the left side.

$n = \pm \sqrt[4]{50625}$

Simplify $\pm \sqrt[4]{50625}$ .

Rewrite $50625$ as $15^{4}$ .

$n = \pm \sqrt[4]{15^{4}}$

Pull terms out from under the radical, assuming positive real numbers.

$n = \pm 15$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$n = 15$

Next, use the negative value of the $\pm$ to find the second solution.

$n = - 15$

The complete solution is the result of both the positive and negative portions of the solution.

$n = 15, - 15$

Exclude the solutions that do not make $\log_{4} (n) = \frac{1}{4} \cdot \log_{4} (81) + \frac{1}{2} \cdot \log_{4} (25)$ true.

$n = 15$

Solve for n log base 4 of n=1/4 log base 4 of 81+1/2 log base 4 of 25

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