Precalculus Examples

$\frac{1}{2} \cdot \log_{12} (9) + \frac{1}{3} \cdot \log_{12} (7)$

Step 1

Simplify each term.

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Step 1.1

Simplify $\frac{1}{2} \log_{12} (9)$ by moving $\frac{1}{2}$ inside the logarithm.

$\log_{12} (9^{\frac{1}{2}}) + \frac{1}{3} \cdot \log_{12} (7)$

Step 1.2

Rewrite $9$ as $3^{2}$ .

$\log_{12} ({(3^{2})}^{\frac{1}{2}}) + \frac{1}{3} \cdot \log_{12} (7)$

Step 1.3

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

$\log_{12} (3^{2 (\frac{1}{2})}) + \frac{1}{3} \cdot \log_{12} (7)$

Step 1.4

Cancel the common factor of $2$ .

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Step 1.4.1

Cancel the common factor.

$\log_{12} (3^{2 (\frac{1}{2})}) + \frac{1}{3} \cdot \log_{12} (7)$

Step 1.4.2

Rewrite the expression.

$\log_{12} (3^{1}) + \frac{1}{3} \cdot \log_{12} (7)$

Step 1.5

Evaluate the exponent.

$\log_{12} (3) + \frac{1}{3} \cdot \log_{12} (7)$

Step 1.6

Simplify $\frac{1}{3} \log_{12} (7)$ by moving $\frac{1}{3}$ inside the logarithm.

$\log_{12} (3) + \log_{12} (7^{\frac{1}{3}})$

Step 2

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

$\log_{12} (3 \cdot 7^{\frac{1}{3}})$

Step 3

Rewrite $\log_{12} (3 \cdot 7^{\frac{1}{3}})$ using the change of base formula.

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Step 3.1

The change of base rule can be used if $a$ and $b$ are greater than $0$ and not equal to $1$ , and $x$ is greater than $0$ .

$\log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)}$

Step 3.2

Substitute in values for the variables in the change of base formula, using $b = 10$ .

$\frac{\log (3 \cdot 7^{\frac{1}{3}})}{\log (12)}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$\frac{\log (3 \cdot 7^{\frac{1}{3}})}{\log (12)}$

Decimal Form:

$0.70314472 \dots$