Question

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Answer 1

$\log (\sqrt[5]{108})$

According to $\sqrt[m]{a}\ \ \iff\ \ a^{^{\frac{1}{m}}}$, we write,

$\log (\sqrt[5]{108})\ \ \ \ \Longrightarrow\ \ \ \log (108^{^{\frac{1}{5}}})$

And we remember $\log (a^b)\ \ \ \iff\ \ \ b\log (a).$ So,

$\log (108^{^{\frac{1}{5}}})\ \ \ \Longrightarrow\ \ \ \dfrac{1}{5}\log (108)$

Now let's write 108 in terms of 2 and 3. Just prime factorization is needed. Skipping the method...

108 = 2² × 3³

So,

$\dfrac{1}{5}\log (108)\ \ \ \Longrightarrow\ \ \ \dfrac{\log (2^2\times 3^3)}{5}$

And we use $\log(a\times b)=\log(a)+\log(b).$ So,

$\dfrac{\log (2^2\times 3^3)}{5}\ \ \ \Longrightarrow\ \ \ \dfrac{\log (2^2)+\log(3^3)}{5}\\ \\ \\ \dfrac{\log (2^2)+\log(3^3)}{5}\ \ \ \Longrightarrow\ \ \ \dfrac{2\log (2)+3\log(3)}{5}\\ \\ \\ \Longrightarrow\ \dfrac{2}{5}\log(2)+\dfrac{3}{5}\log(3)$

Hence expressed!

Here the base of each logarithm is 10. It's not identified and there's no need to identify since it's 10.