Precalculus Examples

$- \log (4.23) + \log (10^{- 13})$

Step 1

Simplify each term.

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Step 1.1

Log base $10$ of $4.23$ is approximately $0.62634036$ .

$- 1 \cdot 0.62634036 + \log (10^{- 13})$

Step 1.2

Multiply $- 1$ by $0.62634036$ .

$- 0.62634036 + \log (10^{- 13})$

Step 1.3

Use logarithm rules to move $- 13$ out of the exponent.

$- 0.62634036 - 13 \log (10)$

Step 1.4

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

$- 0.62634036 - 13 \cdot 1$

Step 1.5

Multiply $- 13$ by $1$ .

$- 0.62634036 - 13$

Step 2

Subtract $13$ from $- 0.62634036$ .

$- 13.62634036$