Precalculus Examples

$\log ({(2)}^{6}) - \log ({(2)}^{15}) + \log ({(2)}^{20})$

Step 1

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

$\log (\frac{{(2)}^{6}}{{(2)}^{15}}) + \log ({(2)}^{20})$

Step 2

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

$\log (\frac{{(2)}^{6}}{{(2)}^{15}} {(2)}^{20})$

Step 3

Cancel the common factor of ${(2)}^{15}$ .

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Step 3.1

Factor $2^{15}$ out of ${(2)}^{20}$ .

$\log (\frac{{(2)}^{6}}{{(2)}^{15}} (2^{15} \cdot 2^{5}))$

Step 3.2

Cancel the common factor.

$\log (\frac{{(2)}^{6}}{{(2)}^{15}} (2^{15} \cdot 2^{5}))$

Step 3.3

Rewrite the expression.

$\log ({(2)}^{6} \cdot 2^{5})$

Step 4

Multiply ${(2)}^{6}$ by $2^{5}$ by adding the exponents.

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Step 4.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\log (2^{6 + 5})$

Step 4.2

Add $6$ and $5$ .

$\log (2^{11})$

Step 5

Raise $2$ to the power of $11$ .

$\log (2048)$

Step 6

The result can be shown in multiple forms.

Exact Form:

$\log (2048)$

Decimal Form:

$3.31132995 \dots$