Precalculus Examples

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

Step 1

Expand $(\log_{2} (3) + \log_{4} (9)) (\log_{3} (2) + \log_{9} (4))$ using the FOIL Method.

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Step 1.1

Apply the distributive property.

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

Step 1.2

Apply the distributive property.

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

Step 1.3

Apply the distributive property.

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

Step 2

Simplify each term.

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Step 2.1

Rewrite $\log_{2} (3)$ using the change of base formula.

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Step 2.1.1

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 $0$ .

$\log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)} \log_{3} (2) + \log_{2} (3) \log_{9} (4) + \log_{4} (9) \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.1.2

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

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

Step 2.2

Rewrite $\log_{3} (2)$ using the change of base formula.

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Step 2.2.1

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 $0$ .

$\frac{\log (3)}{\log (2)} \log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)} + \log_{2} (3) \log_{9} (4) + \log_{4} (9) \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.2.2

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

$\frac{\log (3)}{\log (2)} \cdot \frac{\log (2)}{\log (3)} + \log_{2} (3) \log_{9} (4) + \log_{4} (9) \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.3

Cancel the common factor of $\log (3)$ .

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Step 2.3.1

Cancel the common factor.

$\frac{\log (3)}{\log (2)} \cdot \frac{\log (2)}{\log (3)} + \log_{2} (3) \log_{9} (4) + \log_{4} (9) \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.3.2

Rewrite the expression.

$\frac{1}{\log (2)} \log (2) + \log_{2} (3) \log_{9} (4) + \log_{4} (9) \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.4

Cancel the common factor of $\log (2)$ .

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Step 2.4.1

Cancel the common factor.

$\frac{1}{\log (2)} \log (2) + \log_{2} (3) \log_{9} (4) + \log_{4} (9) \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.4.2

Rewrite the expression.

$1 + \log_{2} (3) \log_{9} (4) + \log_{4} (9) \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.5

Rewrite $\log_{2} (3)$ using the change of base formula.

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Step 2.5.1

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 $0$ .

$1 + \log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)} \log_{9} (4) + \log_{4} (9) \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.5.2

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

$1 + \frac{\log (3)}{\log (2)} \log_{9} (4) + \log_{4} (9) \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.6

Rewrite $\log_{9} (4)$ using the change of base formula.

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Step 2.6.1

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 $0$ .

$1 + \frac{\log (3)}{\log (2)} \log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)} + \log_{4} (9) \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.6.2

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

$1 + \frac{\log (3)}{\log (2)} \cdot \frac{\log (4)}{\log (9)} + \log_{4} (9) \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.7

Rewrite $\log (4)$ as $\log (2^{2})$ .

$1 + \frac{\log (3)}{\log (2)} \cdot \frac{\log (2^{2})}{\log (9)} + \log_{4} (9) \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.8

Expand $\log (2^{2})$ by moving $2$ outside the logarithm.

$1 + \frac{\log (3)}{\log (2)} \cdot \frac{2 \log (2)}{\log (9)} + \log_{4} (9) \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.9

Rewrite $\log (9)$ as $\log (3^{2})$ .

$1 + \frac{\log (3)}{\log (2)} \cdot \frac{2 \log (2)}{\log (3^{2})} + \log_{4} (9) \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.10

Expand $\log (3^{2})$ by moving $2$ outside the logarithm.

$1 + \frac{\log (3)}{\log (2)} \cdot \frac{2 \log (2)}{2 \log (3)} + \log_{4} (9) \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.11

Cancel the common factor of $2$ .

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Step 2.11.1

Cancel the common factor.

$1 + \frac{\log (3)}{\log (2)} \cdot \frac{2 \log (2)}{2 \log (3)} + \log_{4} (9) \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.11.2

Rewrite the expression.

$1 + \frac{\log (3)}{\log (2)} \cdot \frac{\log (2)}{\log (3)} + \log_{4} (9) \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.12

Cancel the common factor of $\log (3)$ .

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Step 2.12.1

Cancel the common factor.

$1 + \frac{\log (3)}{\log (2)} \cdot \frac{\log (2)}{\log (3)} + \log_{4} (9) \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.12.2

Rewrite the expression.

$1 + \frac{1}{\log (2)} \log (2) + \log_{4} (9) \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.13

Cancel the common factor of $\log (2)$ .

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Step 2.13.1

Cancel the common factor.

$1 + \frac{1}{\log (2)} \log (2) + \log_{4} (9) \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.13.2

Rewrite the expression.

$1 + 1 + \log_{4} (9) \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.14

Rewrite $\log_{4} (9)$ using the change of base formula.

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Step 2.14.1

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 $0$ .

$1 + 1 + \log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)} \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.14.2

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

$1 + 1 + \frac{\log (9)}{\log (4)} \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.15

Rewrite $\log (9)$ as $\log (3^{2})$ .

$1 + 1 + \frac{\log (3^{2})}{\log (4)} \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.16

Expand $\log (3^{2})$ by moving $2$ outside the logarithm.

$1 + 1 + \frac{2 \log (3)}{\log (4)} \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.17

Rewrite $\log (4)$ as $\log (2^{2})$ .

$1 + 1 + \frac{2 \log (3)}{\log (2^{2})} \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.18

Expand $\log (2^{2})$ by moving $2$ outside the logarithm.

$1 + 1 + \frac{2 \log (3)}{2 \log (2)} \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.19

Cancel the common factor of $2$ .

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Step 2.19.1

Cancel the common factor.

$1 + 1 + \frac{2 \log (3)}{2 \log (2)} \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.19.2

Rewrite the expression.

$1 + 1 + \frac{\log (3)}{\log (2)} \log_{3} (2) + \log_{4} (9) \log_{9} (4)$

Step 2.20

Rewrite $\log_{3} (2)$ using the change of base formula.

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Step 2.20.1

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 $0$ .

$1 + 1 + \frac{\log (3)}{\log (2)} \log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)} + \log_{4} (9) \log_{9} (4)$

Step 2.20.2

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

$1 + 1 + \frac{\log (3)}{\log (2)} \cdot \frac{\log (2)}{\log (3)} + \log_{4} (9) \log_{9} (4)$

Step 2.21

Cancel the common factor of $\log (3)$ .

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Step 2.21.1

Cancel the common factor.

$1 + 1 + \frac{\log (3)}{\log (2)} \cdot \frac{\log (2)}{\log (3)} + \log_{4} (9) \log_{9} (4)$

Step 2.21.2

Rewrite the expression.

$1 + 1 + \frac{1}{\log (2)} \log (2) + \log_{4} (9) \log_{9} (4)$

Step 2.22

Cancel the common factor of $\log (2)$ .

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Step 2.22.1

Cancel the common factor.

$1 + 1 + \frac{1}{\log (2)} \log (2) + \log_{4} (9) \log_{9} (4)$

Step 2.22.2

Rewrite the expression.

$1 + 1 + 1 + \log_{4} (9) \log_{9} (4)$

Step 2.23

Rewrite $\log_{4} (9)$ using the change of base formula.

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Step 2.23.1

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 $0$ .

$1 + 1 + 1 + \log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)} \log_{9} (4)$

Step 2.23.2

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

$1 + 1 + 1 + \frac{\log (9)}{\log (4)} \log_{9} (4)$

Step 2.24

Rewrite $\log (9)$ as $\log (3^{2})$ .

$1 + 1 + 1 + \frac{\log (3^{2})}{\log (4)} \log_{9} (4)$

Step 2.25

Expand $\log (3^{2})$ by moving $2$ outside the logarithm.

$1 + 1 + 1 + \frac{2 \log (3)}{\log (4)} \log_{9} (4)$

Step 2.26

Rewrite $\log (4)$ as $\log (2^{2})$ .

$1 + 1 + 1 + \frac{2 \log (3)}{\log (2^{2})} \log_{9} (4)$

Step 2.27

Expand $\log (2^{2})$ by moving $2$ outside the logarithm.

$1 + 1 + 1 + \frac{2 \log (3)}{2 \log (2)} \log_{9} (4)$

Step 2.28

Cancel the common factor of $2$ .

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Step 2.28.1

Cancel the common factor.

$1 + 1 + 1 + \frac{2 \log (3)}{2 \log (2)} \log_{9} (4)$

Step 2.28.2

Rewrite the expression.

$1 + 1 + 1 + \frac{\log (3)}{\log (2)} \log_{9} (4)$

Step 2.29

Rewrite $\log_{9} (4)$ using the change of base formula.

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Step 2.29.1

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 $0$ .

$1 + 1 + 1 + \frac{\log (3)}{\log (2)} \log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)}$

Step 2.29.2

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

$1 + 1 + 1 + \frac{\log (3)}{\log (2)} \cdot \frac{\log (4)}{\log (9)}$

Step 2.30

Rewrite $\log (4)$ as $\log (2^{2})$ .

$1 + 1 + 1 + \frac{\log (3)}{\log (2)} \cdot \frac{\log (2^{2})}{\log (9)}$

Step 2.31

Expand $\log (2^{2})$ by moving $2$ outside the logarithm.

$1 + 1 + 1 + \frac{\log (3)}{\log (2)} \cdot \frac{2 \log (2)}{\log (9)}$

Step 2.32

Rewrite $\log (9)$ as $\log (3^{2})$ .

$1 + 1 + 1 + \frac{\log (3)}{\log (2)} \cdot \frac{2 \log (2)}{\log (3^{2})}$

Step 2.33

Expand $\log (3^{2})$ by moving $2$ outside the logarithm.

$1 + 1 + 1 + \frac{\log (3)}{\log (2)} \cdot \frac{2 \log (2)}{2 \log (3)}$

Step 2.34

Cancel the common factor of $2$ .

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Step 2.34.1

Cancel the common factor.

$1 + 1 + 1 + \frac{\log (3)}{\log (2)} \cdot \frac{2 \log (2)}{2 \log (3)}$

Step 2.34.2

Rewrite the expression.

$1 + 1 + 1 + \frac{\log (3)}{\log (2)} \cdot \frac{\log (2)}{\log (3)}$

Step 2.35

Cancel the common factor of $\log (3)$ .

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Step 2.35.1

Cancel the common factor.

$1 + 1 + 1 + \frac{\log (3)}{\log (2)} \cdot \frac{\log (2)}{\log (3)}$

Step 2.35.2

Rewrite the expression.

$1 + 1 + 1 + \frac{1}{\log (2)} \log (2)$

Step 2.36

Cancel the common factor of $\log (2)$ .

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Step 2.36.1

Cancel the common factor.

$1 + 1 + 1 + \frac{1}{\log (2)} \log (2)$

Step 2.36.2

Rewrite the expression.

$1 + 1 + 1 + 1$

Step 3

Simplify by adding numbers.

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Step 3.1

Add $1$ and $1$ .

$2 + 1 + 1$

Step 3.2

Add $2$ and $1$ .

$3 + 1$

Step 3.3

Add $3$ and $1$ .

$4$