Precalculus Examples

$\log_{42} (7^{49}) + \log_{42} (6^{49})$

Step 1

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log_{42} (7^{49} \cdot 6^{49})$

Step 2

Rewrite $\log_{42} (7^{49} \cdot 6^{49})$ using the change of base formula.

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Step 2.1

The change of base rule can be used if $a$ and $b$ are greater than $0$ and not equal to $1$ , and $x$ is greater than $0$ .

$\log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)}$

Step 2.2

Substitute in values for the variables in the change of base formula, using $b = 10$ .

$\frac{\log (7^{49} \cdot 6^{49})}{\log (42)}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$\frac{\log (7^{49} \cdot 6^{49})}{\log (42)}$

Decimal Form:

$49$