Precalculus Examples

$\log_{2} (72) - \frac{1}{2} \cdot (\log_{2} (3) + \log_{2} (27))$

Step 1

Simplify each term.

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Step 1.1

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

$\log_{2} (72) - \frac{1}{2} \cdot \log_{2} (3 \cdot 27)$

Step 1.2

Multiply $3$ by $27$ .

$\log_{2} (72) - \frac{1}{2} \cdot \log_{2} (81)$

Step 1.3

Multiply $- \frac{1}{2} \log_{2} (81)$ .

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Step 1.3.1

Reorder $\log_{2} (81)$ and $\frac{1}{2}$ .

$\log_{2} (72) - (\frac{1}{2} \log_{2} (81))$

Step 1.3.2

Simplify $\frac{1}{2} \log_{2} (81)$ by moving $\frac{1}{2}$ inside the logarithm.

$\log_{2} (72) - \log_{2} (81^{\frac{1}{2}})$

Step 1.4

Rewrite $81$ as $9^{2}$ .

$\log_{2} (72) - \log_{2} ({(9^{2})}^{\frac{1}{2}})$

Step 1.5

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

$\log_{2} (72) - \log_{2} (9^{2 (\frac{1}{2})})$

Step 1.6

Cancel the common factor of $2$ .

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Step 1.6.1

Cancel the common factor.

$\log_{2} (72) - \log_{2} (9^{2 (\frac{1}{2})})$

Step 1.6.2

Rewrite the expression.

$\log_{2} (72) - \log_{2} (9^{1})$

Step 1.7

Evaluate the exponent.

$\log_{2} (72) - \log_{2} (9)$

Step 2

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

$\log_{2} (\frac{72}{9})$

Step 3

Logarithm base $2$ of $\frac{72}{9}$ is $3$ .

$3$