Finite Math Examples

$t = \frac{\log (15300) - \log (8000)}{4 \log (1.015)}$

Step 1

Simplify the numerator.

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Step 1.1

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

$t = \frac{\log (\frac{15300}{8000})}{4 \log (1.015)}$

Step 1.2

Cancel the common factor of $15300$ and $8000$ .

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Step 1.2.1

Factor $100$ out of $15300$ .

$t = \frac{\log (\frac{100 (153)}{8000})}{4 \log (1.015)}$

Step 1.2.2

Cancel the common factors.

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Step 1.2.2.1

Factor $100$ out of $8000$ .

$t = \frac{\log (\frac{100 \cdot 153}{100 \cdot 80})}{4 \log (1.015)}$

Step 1.2.2.2

Cancel the common factor.

$t = \frac{\log (\frac{100 \cdot 153}{100 \cdot 80})}{4 \log (1.015)}$

Step 1.2.2.3

Rewrite the expression.

$t = \frac{\log (\frac{153}{80})}{4 \log (1.015)}$

Step 2

Simplify $4 \log (1.015)$ by moving $4$ inside the logarithm.

$t = \frac{\log (\frac{153}{80})}{\log ({1.015}^{4})}$

Step 3

Simplify the denominator.

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Step 3.1

Raise $1.015$ to the power of $4$ .

$t = \frac{\log (\frac{153}{80})}{\log (1.06136355)}$

Step 3.2

Log base $10$ of $1.06136355$ is approximately $0.02586416$ .

$t = \frac{\log (\frac{153}{80})}{0.02586416}$

Step 4

Divide $153$ by $80$ .

$t = \frac{\log (1.9125)}{0.02586416}$

Step 5

Log base $10$ of $1.9125$ is approximately $0.28160144$ .

$t = \frac{0.28160144}{0.02586416}$

Step 6

Divide $0.28160144$ by $0.02586416$ .

$t = 10.88770506$