Precalculus Examples

$\frac{2}{3} \cdot \log_{72} (27) + \frac{3}{4} \cdot \log_{72} (16)$

Step 1

Simplify each term.

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Step 1.1

Simplify $\frac{2}{3} \log_{72} (27)$ by moving $\frac{2}{3}$ inside the logarithm.

$\log_{72} (27^{\frac{2}{3}}) + \frac{3}{4} \cdot \log_{72} (16)$

Step 1.2

Rewrite $27$ as $3^{3}$ .

$\log_{72} ({(3^{3})}^{\frac{2}{3}}) + \frac{3}{4} \cdot \log_{72} (16)$

Step 1.3

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

$\log_{72} (3^{3 (\frac{2}{3})}) + \frac{3}{4} \cdot \log_{72} (16)$

Step 1.4

Cancel the common factor of $3$ .

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Step 1.4.1

Cancel the common factor.

$\log_{72} (3^{3 (\frac{2}{3})}) + \frac{3}{4} \cdot \log_{72} (16)$

Step 1.4.2

Rewrite the expression.

$\log_{72} (3^{2}) + \frac{3}{4} \cdot \log_{72} (16)$

Step 1.5

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

$\log_{72} (9) + \frac{3}{4} \cdot \log_{72} (16)$

Step 1.6

Simplify $\frac{3}{4} \log_{72} (16)$ by moving $\frac{3}{4}$ inside the logarithm.

$\log_{72} (9) + \log_{72} (16^{\frac{3}{4}})$

Step 1.7

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

$\log_{72} (9) + \log_{72} ({(2^{4})}^{\frac{3}{4}})$

Step 1.8

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

$\log_{72} (9) + \log_{72} (2^{4 (\frac{3}{4})})$

Step 1.9

Cancel the common factor of $4$ .

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Step 1.9.1

Cancel the common factor.

$\log_{72} (9) + \log_{72} (2^{4 (\frac{3}{4})})$

Step 1.9.2

Rewrite the expression.

$\log_{72} (9) + \log_{72} (2^{3})$

Step 1.10

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

$\log_{72} (9) + \log_{72} (8)$

Step 2

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

$\log_{72} (9 \cdot 8)$

Step 3

Multiply $9$ by $8$ .

$\log_{72} (72)$

Step 4

Logarithm base $72$ of $72$ is $1$ .

$1$