Precalculus Examples

$\log_{5} (6.25) \cdot \log_{5} (100) \cdot \log_{5} (25)$

Step 1

Log base $5$ of $6.25$ is approximately $1.13864688$ .

$1.13864688 \cdot \log_{5} (100) \cdot \log_{5} (25)$

Step 2

Simplify $1.13864688 \log_{5} (100)$ by moving $1.13864688$ inside the logarithm.

$\log_{5} (100^{1.13864688}) \cdot \log_{5} (25)$

Step 3

Raise $100$ to the power of $1.13864688$ .

$\log_{5} (189.36240804) \cdot \log_{5} (25)$

Step 4

Log base $5$ of $189.36240804$ is approximately $3.2580708$ .

$3.2580708 \cdot \log_{5} (25)$

Step 5

Logarithm base $5$ of $25$ is $2$ .

$3.2580708 \cdot 2$

Step 6

Multiply $3.2580708$ by $2$ .

$6.51614161$