Evaluate 3 log base 7 of sixth root of 49

$3 \log_{7} (\sqrt[6]{49})$

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

$3 \log_{7} (\sqrt[6]{7^{2}})$

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

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

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

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

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

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

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

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{7}$ as $7^{\frac{1}{3}}$ .

$\log_{7} ({(7^{\frac{1}{3}})}^{3})$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Combine $\frac{1}{3}$ and $3$ .

$\log_{7} (7^{\frac{3}{3}})$

Cancel the common factor of $3$ .

Cancel the common factor.

$\log_{7} (7^{\frac{3}{3}})$

Rewrite the expression.

$\log_{7} (7^{1})$

Evaluate the exponent.

$\log_{7} (7)$

Logarithm base $7$ of $7$ is $1$ .

$1$