Precalculus Examples

$\log_{3} (y) = \frac{1}{4} \cdot \log_{3} (16) + \frac{1}{3} \cdot \log_{3} (64)$

Step 1

Simplify the right side.

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Step 1.1

Simplify each term.

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Step 1.1.1

Combine $\frac{1}{4}$ and $\log_{3} (16)$ .

$\log_{3} (y) = \frac{\log_{3} (16)}{4} + \frac{1}{3} \log_{3} (64)$

Step 1.1.2

Combine $\frac{1}{3}$ and $\log_{3} (64)$ .

$\log_{3} (y) = \frac{\log_{3} (16)}{4} + \frac{\log_{3} (64)}{3}$

Step 2

Multiply each term in $\log_{3} (y) = \frac{\log_{3} (16)}{4} + \frac{\log_{3} (64)}{3}$ by $12$ to eliminate the fractions.

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Step 2.1

Multiply each term in $\log_{3} (y) = \frac{\log_{3} (16)}{4} + \frac{\log_{3} (64)}{3}$ by $12$ .

$\log_{3} (y) \cdot 12 = \frac{\log_{3} (16)}{4} \cdot 12 + \frac{\log_{3} (64)}{3} \cdot 12$

Step 2.2

Simplify the left side.

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Step 2.2.1

Move $12$ to the left of $\log_{3} (y)$ .

$12 \log_{3} (y) = \frac{\log_{3} (16)}{4} \cdot 12 + \frac{\log_{3} (64)}{3} \cdot 12$

Step 2.3

Simplify the right side.

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Step 2.3.1

Simplify each term.

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Step 2.3.1.1

Cancel the common factor of $4$ .

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Step 2.3.1.1.1

Factor $4$ out of $12$ .

$12 \log_{3} (y) = \frac{\log_{3} (16)}{4} \cdot (4 (3)) + \frac{\log_{3} (64)}{3} \cdot 12$

Step 2.3.1.1.2

Cancel the common factor.

$12 \log_{3} (y) = \frac{\log_{3} (16)}{4} \cdot (4 \cdot 3) + \frac{\log_{3} (64)}{3} \cdot 12$

Step 2.3.1.1.3

Rewrite the expression.

$12 \log_{3} (y) = \log_{3} (16) \cdot 3 + \frac{\log_{3} (64)}{3} \cdot 12$

Step 2.3.1.2

Move $3$ to the left of $\log_{3} (16)$ .

$12 \log_{3} (y) = 3 \cdot \log_{3} (16) + \frac{\log_{3} (64)}{3} \cdot 12$

Step 2.3.1.3

Cancel the common factor of $3$ .

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Step 2.3.1.3.1

Factor $3$ out of $12$ .

$12 \log_{3} (y) = 3 \log_{3} (16) + \frac{\log_{3} (64)}{3} \cdot (3 (4))$

Step 2.3.1.3.2

Cancel the common factor.

$12 \log_{3} (y) = 3 \log_{3} (16) + \frac{\log_{3} (64)}{3} \cdot (3 \cdot 4)$

Step 2.3.1.3.3

Rewrite the expression.

$12 \log_{3} (y) = 3 \log_{3} (16) + \log_{3} (64) \cdot 4$

Step 2.3.1.4

Move $4$ to the left of $\log_{3} (64)$ .

$12 \log_{3} (y) = 3 \log_{3} (16) + 4 \log_{3} (64)$

Step 3

Simplify the left side.

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Step 3.1

Simplify $12 \log_{3} (y)$ by moving $12$ inside the logarithm.

$\log_{3} (y^{12}) = 3 \log_{3} (16) + 4 \log_{3} (64)$

Step 4

Simplify the right side.

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Step 4.1

Simplify $3 \log_{3} (16) + 4 \log_{3} (64)$ .

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Step 4.1.1

Simplify each term.

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Step 4.1.1.1

Simplify $3 \log_{3} (16)$ by moving $3$ inside the logarithm.

$\log_{3} (y^{12}) = \log_{3} (16^{3}) + 4 \log_{3} (64)$

Step 4.1.1.2

Raise $16$ to the power of $3$ .

$\log_{3} (y^{12}) = \log_{3} (4096) + 4 \log_{3} (64)$

Step 4.1.1.3

Simplify $4 \log_{3} (64)$ by moving $4$ inside the logarithm.

$\log_{3} (y^{12}) = \log_{3} (4096) + \log_{3} (64^{4})$

Step 4.1.1.4

Raise $64$ to the power of $4$ .

$\log_{3} (y^{12}) = \log_{3} (4096) + \log_{3} (16777216)$

Step 4.1.2

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

$\log_{3} (y^{12}) = \log_{3} (4096 \cdot 16777216)$

Step 4.1.3

Multiply $4096$ by $16777216$ .

$\log_{3} (y^{12}) = \log_{3} (68719476736)$

Step 5

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

$y^{12} = 68719476736$

Step 6

Solve for $y$ .

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Step 6.1

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

$y = \pm \sqrt[12]{68719476736}$

Step 6.2

Simplify $\pm \sqrt[12]{68719476736}$ .

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Step 6.2.1

Rewrite $68719476736$ as $8^{12}$ .

$y = \pm \sqrt[12]{8^{12}}$

Step 6.2.2

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

$y = \pm 8$

Step 6.3

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

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Step 6.3.1

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

$y = 8$

Step 6.3.2

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

$y = - 8$

Step 6.3.3

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

$y = 8, - 8$

Step 7

Exclude the solutions that do not make $\log_{3} (y) = \frac{1}{4} \cdot \log_{3} (16) + \frac{1}{3} \cdot \log_{3} (64)$ true.

$y = 8$