Precalculus Examples

$3 \log_{7} (2) + \frac{1}{2} \cdot \log_{7} (49)$

Step 1

Simplify $3 \log_{7} (2)$ by moving $3$ inside the logarithm.

$\log_{7} (2^{3}) + \frac{1}{2} \cdot \log_{7} (49)$

Step 2

Raise $2$ to the power of $3$ .

$\log_{7} (8) + \frac{1}{2} \cdot \log_{7} (49)$

Step 3

Logarithm base $7$ of $49$ is $2$ .

$\log_{7} (8) + \frac{1}{2} \cdot 2$

Step 4

Cancel the common factor of $2$ .

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Step 4.1

Cancel the common factor.

$\log_{7} (8) + \frac{1}{2} \cdot 2$

Step 4.2

Rewrite the expression.

$\log_{7} (8) + 1$

Step 5

The result can be shown in multiple forms.

Exact Form:

$\log_{7} (8) + 1$

Decimal Form:

$2.06862156 \dots$