Precalculus Examples

$\log_{4} (24) + \log_{4} (\frac{64}{3}) - \log_{4} (32)$

Step 1

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

$\log_{4} (24 (\frac{64}{3})) - \log_{4} (32)$

Step 2

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

$\log_{4} (\frac{24 (\frac{64}{3})}{32})$

Step 3

Reduce the expression $\frac{24 (\frac{64}{3})}{32}$ by cancelling the common factors.

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Step 3.1

Factor $8$ out of $24 (\frac{64}{3})$ .

$\log_{4} (\frac{8 (3 (\frac{64}{3}))}{32})$

Step 3.2

Factor $8$ out of $32$ .

$\log_{4} (\frac{8 (3 (\frac{64}{3}))}{8 (4)})$

Step 3.3

Cancel the common factor.

$\log_{4} (\frac{8 (3 (\frac{64}{3}))}{8 \cdot 4})$

Step 3.4

Rewrite the expression.

$\log_{4} (\frac{3 (\frac{64}{3})}{4})$

Step 4

Rewrite $\frac{3 (\frac{64}{3})}{4}$ as $3 (\frac{64}{3}) \cdot 4^{- 1}$ .

$\log_{4} (3 (\frac{64}{3}) \cdot 4^{- 1})$

Step 5

Rewrite $\log_{4} (3 (\frac{64}{3}) \cdot 4^{- 1})$ as $\log_{4} (3 (\frac{64}{3})) + \log_{4} (4^{- 1})$ .

$\log_{4} (3 (\frac{64}{3})) + \log_{4} (4^{- 1})$

Step 6

Use logarithm rules to move $- 1$ out of the exponent.

$\log_{4} (3 (\frac{64}{3})) - \log_{4} (4)$

Step 7

Logarithm base $4$ of $4$ is $1$ .

$\log_{4} (3 (\frac{64}{3})) - 1 \cdot 1$

Step 8

Multiply $- 1$ by $1$ .

$\log_{4} (3 (\frac{64}{3})) - 1$

Step 9

Rewrite $\log_{4} (3 (\frac{64}{3}))$ as $\log_{4} (3) + \log_{4} (\frac{64}{3})$ .

$\log_{4} (3) + \log_{4} (\frac{64}{3}) - 1$

Step 10

The result can be shown in multiple forms.

Exact Form:

$\log_{4} (3) + \log_{4} (\frac{64}{3}) - 1$

Decimal Form:

$2$