Precalculus Examples

$\frac{\log_{3} (\frac{3}{4}) + \log_{3} (\frac{4}{5}) + \log_{3} (\frac{5}{6}) + \log_{3} (\frac{6}{7}) + \log_{3} (\frac{7}{8}) + \log_{3} (\frac{8}{9})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1

Simplify the numerator.

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Step 1.1

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

$\frac{\log_{3} (\frac{3}{4} \cdot \frac{4}{5}) + \log_{3} (\frac{5}{6}) + \log_{3} (\frac{6}{7}) + \log_{3} (\frac{7}{8}) + \log_{3} (\frac{8}{9})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1.2

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

$\frac{\log_{3} (\frac{3}{4} \cdot \frac{4}{5} \frac{5}{6}) + \log_{3} (\frac{6}{7}) + \log_{3} (\frac{7}{8}) + \log_{3} (\frac{8}{9})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1.3

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

$\frac{\log_{3} (\frac{3}{4} \cdot \frac{4}{5} \frac{5}{6} \frac{6}{7}) + \log_{3} (\frac{7}{8}) + \log_{3} (\frac{8}{9})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1.4

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

$\frac{\log_{3} (\frac{3}{4} \cdot \frac{4}{5} \frac{5}{6} \frac{6}{7} \frac{7}{8}) + \log_{3} (\frac{8}{9})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1.5

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

$\frac{\log_{3} (\frac{3}{4} \cdot \frac{4}{5} \frac{5}{6} \frac{6}{7} \frac{7}{8} \frac{8}{9})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1.6

Cancel the common factor of $4$ .

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Step 1.6.1

Cancel the common factor.

Step 1.6.2

Rewrite the expression.

$\frac{\log_{3} (3 (\frac{1}{5}) \frac{5}{6} \frac{6}{7} \frac{7}{8} \frac{8}{9})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1.7

Combine $3$ and $\frac{1}{5}$ .

$\frac{\log_{3} (\frac{3}{5} \cdot \frac{5}{6} \frac{6}{7} \frac{7}{8} \frac{8}{9})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1.8

Cancel the common factor of $3$ .

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Step 1.8.1

Factor $3$ out of $6$ .

$\frac{\log_{3} (\frac{3}{5} \cdot \frac{5}{3 (2)} \frac{6}{7} \frac{7}{8} \frac{8}{9})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1.8.2

Cancel the common factor.

$\frac{\log_{3} (\frac{3}{5} \cdot \frac{5}{3 \cdot 2} \frac{6}{7} \frac{7}{8} \frac{8}{9})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1.8.3

Rewrite the expression.

$\frac{\log_{3} (\frac{1}{5} \cdot \frac{5}{2} \frac{6}{7} \frac{7}{8} \frac{8}{9})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1.9

Cancel the common factor of $5$ .

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Step 1.9.1

Cancel the common factor.

Step 1.9.2

Rewrite the expression.

$\frac{\log_{3} (\frac{1}{2} \cdot \frac{6}{7} \frac{7}{8} \frac{8}{9})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1.10

Cancel the common factor of $2$ .

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Step 1.10.1

Factor $2$ out of $6$ .

$\frac{\log_{3} (\frac{1}{2} \cdot \frac{2 (3)}{7} \frac{7}{8} \frac{8}{9})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1.10.2

Cancel the common factor.

$\frac{\log_{3} (\frac{1}{2} \cdot \frac{2 \cdot 3}{7} \frac{7}{8} \frac{8}{9})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1.10.3

Rewrite the expression.

$\frac{\log_{3} (\frac{3}{7} \cdot \frac{7}{8} \frac{8}{9})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1.11

Cancel the common factor of $7$ .

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Step 1.11.1

Cancel the common factor.

$\frac{\log_{3} (\frac{3}{7} \cdot \frac{7}{8} \frac{8}{9})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1.11.2

Rewrite the expression.

$\frac{\log_{3} (3 (\frac{1}{8}) \frac{8}{9})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1.12

Combine $3$ and $\frac{1}{8}$ .

$\frac{\log_{3} (\frac{3}{8} \cdot \frac{8}{9})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1.13

Cancel the common factor of $3$ .

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Step 1.13.1

Factor $3$ out of $9$ .

$\frac{\log_{3} (\frac{3}{8} \cdot \frac{8}{3 (3)})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1.13.2

Cancel the common factor.

$\frac{\log_{3} (\frac{3}{8} \cdot \frac{8}{3 \cdot 3})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1.13.3

Rewrite the expression.

$\frac{\log_{3} (\frac{1}{8} \cdot \frac{8}{3})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1.14

Cancel the common factor of $8$ .

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Step 1.14.1

Cancel the common factor.

$\frac{\log_{3} (\frac{1}{8} \cdot \frac{8}{3})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1.14.2

Rewrite the expression.

$\frac{\log_{3} (\frac{1}{3})}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 1.15

Logarithm base $3$ of $\frac{1}{3}$ is $- 1$ .

$\frac{- 1}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (\frac{9}{10})}$

Step 2

Simplify the denominator.

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Step 2.1

Rewrite $\frac{9}{10}$ as $9 \cdot 10^{- 1}$ .

$\frac{- 1}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (9 \cdot 10^{- 1})}$

Step 2.2

Rewrite $\log (9 \cdot 10^{- 1})$ as $\log (9) + \log (10^{- 1})$ .

$\frac{- 1}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (9) + \log (10^{- 1})}$

Step 2.3

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

$\frac{- 1}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (9) - \log (10)}$

Step 2.4

Logarithm base $10$ of $10$ is $1$ .

$\frac{- 1}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (9) - 1 \cdot 1}$

Step 2.5

Multiply $- 1$ by $1$ .

$\frac{- 1}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (9) - 1}$

Step 2.6

Simplify each term.

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Step 2.6.1

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

$\frac{- 1}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + \log (3^{2}) - 1}$

Step 2.6.2

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

$\frac{- 1}{\log (\frac{4}{5}) + \log (\frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + 2 \log (3) - 1}$

Step 2.7

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

$\frac{- 1}{\log (\frac{4}{5} \cdot \frac{5}{6}) + \log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{8}{9}) + 2 \log (3) - 1}$

Step 2.8

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

$\frac{- 1}{\log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{4}{5} \cdot \frac{5}{6} \frac{8}{9}) + 2 \log (3) - 1}$

Step 2.9

Rewrite $\log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{4}{5} \cdot \frac{5}{6} \frac{8}{9}) + 2 \log (3) - 1$ in a factored form.

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Step 2.9.1

Cancel the common factor of $2$ .

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Step 2.9.1.1

Factor $2$ out of $4$ .

$\frac{- 1}{\log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{2 (2)}{5} \cdot \frac{5}{6} \frac{8}{9}) + 2 \log (3) - 1}$

Step 2.9.1.2

Factor $2$ out of $6$ .

$\frac{- 1}{\log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{2 \cdot 2}{5} \cdot \frac{5}{2 \cdot 3} \frac{8}{9}) + 2 \log (3) - 1}$

Step 2.9.1.3

Cancel the common factor.

$\frac{- 1}{\log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{2 \cdot 2}{5} \cdot \frac{5}{2 \cdot 3} \frac{8}{9}) + 2 \log (3) - 1}$

Step 2.9.1.4

Rewrite the expression.

$\frac{- 1}{\log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{2}{5} \cdot \frac{5}{3} \frac{8}{9}) + 2 \log (3) - 1}$

Step 2.9.2

Cancel the common factor of $5$ .

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Step 2.9.2.1

Cancel the common factor.

$\frac{- 1}{\log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{2}{5} \cdot \frac{5}{3} \frac{8}{9}) + 2 \log (3) - 1}$

Step 2.9.2.2

Rewrite the expression.

$\frac{- 1}{\log (\frac{6}{7}) \log (\frac{7}{8}) + \log (2 (\frac{1}{3}) \frac{8}{9}) + 2 \log (3) - 1}$

Step 2.9.3

Combine $2$ and $\frac{1}{3}$ .

$\frac{- 1}{\log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{2}{3} \cdot \frac{8}{9}) + 2 \log (3) - 1}$

Step 2.9.4

Multiply $\frac{2}{3} \cdot \frac{8}{9}$ .

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Step 2.9.4.1

Multiply $\frac{2}{3}$ by $\frac{8}{9}$ .

$\frac{- 1}{\log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{2 \cdot 8}{3 \cdot 9}) + 2 \log (3) - 1}$

Step 2.9.4.2

Multiply $2$ by $8$ .

$\frac{- 1}{\log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{16}{3 \cdot 9}) + 2 \log (3) - 1}$

Step 2.9.4.3

Multiply $3$ by $9$ .

$\frac{- 1}{\log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{16}{27}) + 2 \log (3) - 1}$

Step 2.9.5

Simplify $2 \log (3)$ by moving $2$ inside the logarithm.

$\frac{- 1}{\log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{16}{27}) + \log (3^{2}) - 1}$

Step 2.9.6

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

$\frac{- 1}{\log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{16}{27}) + \log (9) - 1}$

Step 2.9.7

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

$\frac{- 1}{\log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{16}{27} \cdot 9) - 1}$

Step 2.9.8

Cancel the common factor of $9$ .

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Step 2.9.8.1

Factor $9$ out of $27$ .

$\frac{- 1}{\log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{16}{9 (3)} \cdot 9) - 1}$

Step 2.9.8.2

Cancel the common factor.

$\frac{- 1}{\log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{16}{9 \cdot 3} \cdot 9) - 1}$

Step 2.9.8.3

Rewrite the expression.

$\frac{- 1}{\log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{16}{3}) - 1}$

Step 3

Move the negative in front of the fraction.

$- \frac{1}{\log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{16}{3}) - 1}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$- \frac{1}{\log (\frac{6}{7}) \log (\frac{7}{8}) + \log (\frac{16}{3}) - 1}$

Decimal Form:

$3.71582973 \dots$