Precalculus Examples

$\frac{1}{4} \cdot \log_{b} (16) - 2 \log_{b} (5) + \log_{b} (7)$

Step 1

Simplify each term.

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Step 1.1

Simplify $\frac{1}{4} \log_{b} (16)$ by moving $\frac{1}{4}$ inside the logarithm.

$\log_{b} (16^{\frac{1}{4}}) - 2 \log_{b} (5) + \log_{b} (7)$

Step 1.2

Rewrite $16$ as $2^{4}$ .

$\log_{b} ({(2^{4})}^{\frac{1}{4}}) - 2 \log_{b} (5) + \log_{b} (7)$

Step 1.3

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

$\log_{b} (2^{4 (\frac{1}{4})}) - 2 \log_{b} (5) + \log_{b} (7)$

Step 1.4

Cancel the common factor of $4$ .

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Step 1.4.1

Cancel the common factor.

$\log_{b} (2^{4 (\frac{1}{4})}) - 2 \log_{b} (5) + \log_{b} (7)$

Step 1.4.2

Rewrite the expression.

$\log_{b} (2^{1}) - 2 \log_{b} (5) + \log_{b} (7)$

Step 1.5

Evaluate the exponent.

$\log_{b} (2) - 2 \log_{b} (5) + \log_{b} (7)$

Step 1.6

Simplify $- 2 \log_{b} (5)$ by moving $2$ inside the logarithm.

$\log_{b} (2) - \log_{b} (5^{2}) + \log_{b} (7)$

Step 1.7

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

$\log_{b} (2) - \log_{b} (25) + \log_{b} (7)$

Step 2

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

$\log_{b} (\frac{2}{25}) + \log_{b} (7)$

Step 3

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

$\log_{b} (\frac{2}{25} \cdot 7)$

Step 4

Multiply $\frac{2}{25} \cdot 7$ .

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Step 4.1

Combine $\frac{2}{25}$ and $7$ .

$\log_{b} (\frac{2 \cdot 7}{25})$

Step 4.2

Multiply $2$ by $7$ .

$\log_{b} (\frac{14}{25})$