Precalculus Examples

$- \log (1.6 \cdot 10^{- 2})$

Step 1

Rewrite $\log (1.6 \cdot 10^{- 2})$ as $\log (1.6) + \log (10^{- 2})$ .

$- (\log (1.6) + \log (10^{- 2}))$

Step 2

Expand $\log (10^{- 2})$ by moving $- 2$ outside the logarithm.

$- (\log (1.6) - 2 \log (10))$

Step 3

Log base $10$ of $1.6$ is approximately $0.20411998$ .

$- (0.20411998 - 2 \log (10))$

Step 4

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

$- (0.20411998 - 2 \cdot 1)$

Step 5

Multiply $- 2$ by $1$ .

$- (0.20411998 - 2)$

Step 6

Subtract $2$ from $0.20411998$ .

$- - 1.79588001$

Step 7

Multiply $- 1$ by $- 1.79588001$ .

$1.79588001$