Precalculus Examples

$\frac{1}{5} \cdot \log_{x} (32) - 3 \log_{x} (2) + \log_{x} (4)$

Step 1

Simplify each term.

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Step 1.1

Simplify $\frac{1}{5} \log_{x} (32)$ by moving $\frac{1}{5}$ inside the logarithm.

$\log_{x} (32^{\frac{1}{5}}) - 3 \log_{x} (2) + \log_{x} (4)$

Step 1.2

Rewrite $32$ as $2^{5}$ .

$\log_{x} ({(2^{5})}^{\frac{1}{5}}) - 3 \log_{x} (2) + \log_{x} (4)$

Step 1.3

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\log_{x} (2^{5 (\frac{1}{5})}) - 3 \log_{x} (2) + \log_{x} (4)$

Step 1.4

Cancel the common factor of $5$ .

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Step 1.4.1

Cancel the common factor.

$\log_{x} (2^{5 (\frac{1}{5})}) - 3 \log_{x} (2) + \log_{x} (4)$

Step 1.4.2

Rewrite the expression.

$\log_{x} (2^{1}) - 3 \log_{x} (2) + \log_{x} (4)$

Step 1.5

Evaluate the exponent.

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

Step 1.6

Simplify $- 3 \log_{x} (2)$ by moving $3$ inside the logarithm.

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

Step 1.7

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

$\log_{x} (2) - \log_{x} (8) + \log_{x} (4)$

Step 2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{x} (\frac{2}{8}) + \log_{x} (4)$

Step 3

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

$\log_{x} (\frac{2}{8} \cdot 4)$

Step 4

Cancel the common factor of $4$ .

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Step 4.1

Factor $4$ out of $8$ .

$\log_{x} (\frac{2}{4 (2)} \cdot 4)$

Step 4.2

Cancel the common factor.

$\log_{x} (\frac{2}{4 \cdot 2} \cdot 4)$

Step 4.3

Rewrite the expression.

$\log_{x} (\frac{2}{2})$

Step 5

Divide $2$ by $2$ .

$\log_{x} (1)$

Step 6

Logarithm base $x$ of $1$ is $0$ .

$0$