Precalculus Examples

$\frac{\ln (100)}{\ln (50)} + 1.5$

Step 1

Simplify each term.

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Step 1.1

Simplify the numerator.

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Step 1.1.1

Rewrite $\ln (100)$ as $\ln (2^{2} \cdot 5^{2})$ .

$\frac{\ln (2^{2} \cdot 5^{2})}{\ln (50)} + 1.5$

Step 1.1.2

Rewrite $\ln (2^{2} \cdot 5^{2})$ as $\ln (2^{2}) + \ln (5^{2})$ .

$\frac{\ln (2^{2}) + \ln (5^{2})}{\ln (50)} + 1.5$

Step 1.1.3

Simplify each term.

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Step 1.1.3.1

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

$\frac{2 \ln (2) + \ln (5^{2})}{\ln (50)} + 1.5$

Step 1.1.3.2

Expand $\ln (5^{2})$ by moving $2$ outside the logarithm.

$\frac{2 \ln (2) + 2 \ln (5)}{\ln (50)} + 1.5$

Step 1.1.4

Factor $2$ out of $2 \ln (2) + 2 \ln (5)$ .

$\frac{2 (\ln (2) + \ln (5))}{\ln (50)} + 1.5$

Step 1.2

Simplify the denominator.

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Step 1.2.1

Rewrite $\ln (50)$ as $\ln (2 \cdot 5^{2})$ .

$\frac{2 (\ln (2) + \ln (5))}{\ln (2 \cdot 5^{2})} + 1.5$

Step 1.2.2

Rewrite $\ln (2 \cdot 5^{2})$ as $\ln (2) + \ln (5^{2})$ .

$\frac{2 (\ln (2) + \ln (5))}{\ln (2) + \ln (5^{2})} + 1.5$

Step 1.2.3

Expand $\ln (5^{2})$ by moving $2$ outside the logarithm.

$\frac{2 (\ln (2) + \ln (5))}{\ln (2) + 2 \ln (5)} + 1.5$

Step 2

To write $1.5$ as a fraction with a common denominator, multiply by $\frac{\ln (2) + 2 \ln (5)}{\ln (2) + 2 \ln (5)}$ .

$\frac{2 (\ln (2) + \ln (5))}{\ln (2) + 2 \ln (5)} + \frac{1.5 (\ln (2) + 2 \ln (5))}{\ln (2) + 2 \ln (5)}$

Step 3

Combine the numerators over the common denominator.

$\frac{2 (\ln (2) + \ln (5)) + 1.5 (\ln (2) + 2 \ln (5))}{\ln (2) + 2 \ln (5)}$

Step 4

Simplify the numerator.

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Step 4.1

Apply the distributive property.

$\frac{2 \ln (2) + 2 \ln (5) + 1.5 (\ln (2) + 2 \ln (5))}{\ln (2) + 2 \ln (5)}$

Step 4.2

Multiply $1.5$ by $\ln (2) + 2 \ln (5)$ .

$\frac{2 \ln (2) + 2 \ln (5) + 5.8680345}{\ln (2) + 2 \ln (5)}$

Step 4.3

Add $2 \ln (2)$ and $5.8680345$ .

$\frac{2 \ln (5) + 7.25432886}{\ln (2) + 2 \ln (5)}$

Step 4.4

Add $2 \ln (5)$ and $7.25432886$ .

$\frac{10.47320469}{\ln (2) + 2 \ln (5)}$

Step 5

Replace $e$ with an approximation.

$\frac{10.47320469}{\log_{2.71828182} (2) + 2 \ln (5)}$

Step 6

Log base $2.71828182$ of $2$ is approximately $0.69314718$ .

$\frac{10.47320469}{0.69314718 + 2 \ln (5)}$

Step 7

Replace $e$ with an approximation.

$\frac{10.47320469}{0.69314718 + 2 \log_{2.71828182} (5)}$

Step 8

Log base $2.71828182$ of $5$ is approximately $1.60943791$ .

$\frac{10.47320469}{0.69314718 + 2 \cdot 1.60943791}$

Step 9

Multiply $2$ by $1.60943791$ .

$\frac{10.47320469}{0.69314718 + 3.21887582}$

Step 10

Add $0.69314718$ and $3.21887582$ .

$\frac{10.47320469}{3.912023}$

Step 11

Divide $10.47320469$ by $3.912023$ .

$2.67718382$