Evaluate 5 log base 7 of sixth root of 49

$5 \log_{7} (\sqrt[6]{49})$

Step 1

Rewrite $49$ as $7^{2}$ .

$5 \log_{7} (\sqrt[6]{7^{2}})$

Step 2

Rewrite $\sqrt[6]{7^{2}}$ as $\sqrt[3]{\sqrt{7^{2}}}$ .

$5 \log_{7} (\sqrt[3]{\sqrt{7^{2}}})$

Step 3

Pull terms out from under the radical, assuming positive real numbers.

$5 \log_{7} (\sqrt[3]{7})$

Step 4

Simplify $5 \log_{7} (\sqrt[3]{7})$ by moving $5$ inside the logarithm.

$\log_{7} ({\sqrt[3]{7}}^{5})$

Step 5

Rewrite ${\sqrt[3]{7}}^{5}$ as ${(7^{5})}^{\frac{1}{3}}$ .

$\log_{7} (\sqrt[3]{7^{5}})$

Step 6

Raise $7$ to the power of $5$ .

$\log_{7} (\sqrt[3]{16807})$

Step 7

Rewrite $16807$ as $7^{3} \cdot 49$ .

Factor $343$ out of $16807$ .

$\log_{7} (\sqrt[3]{343 (49)})$

Rewrite $343$ as $7^{3}$ .

$\log_{7} (\sqrt[3]{7^{3} \cdot 49})$

Step 8

Pull terms out from under the radical.

$\log_{7} (7 \sqrt[3]{49})$

Step 9

Rewrite $\log_{7} (7 \sqrt[3]{49})$ using the change of base formula.

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 $0$ .

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

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

$\frac{\log (7 \sqrt[3]{49})}{\log (7)}$

Step 10

The result can be shown in multiple forms.

Exact Form:

$\frac{\log (7 \sqrt[3]{49})}{\log (7)}$

Decimal Form:

$1.\overline{6}$