Precalculus Examples

$\frac{\log (1.133)}{2 \log (1.05)}$

Step 1

Log base $10$ of $1.133$ is approximately $0.0542299$ .

$\frac{0.0542299}{2 \log (1.05)}$

Step 2

Simplify $2 \log (1.05)$ by moving $2$ inside the logarithm.

$\frac{0.0542299}{\log ({1.05}^{2})}$

Step 3

Simplify the denominator.

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Step 3.1

Raise $1.05$ to the power of $2$ .

$\frac{0.0542299}{\log (1.1025)}$

Step 3.2

Log base $10$ of $1.1025$ is approximately $0.04237859$ .

$\frac{0.0542299}{0.04237859}$

Step 4

Divide $0.0542299$ by $0.04237859$ .

$1.27965322$