I'm not great with logarithms so I'd appreciate some help with the following

Question

How is it that $n^{\frac{1}{\log n}} = 10$.

I understand that $10^{\log a} = a$ but I don't know how to make the correct algebraic manipulations.

Note: Assume $log$ is base 10

Answer 1

Hint: $10^{\log n}=n$ by what you're already noted. Now raise both sides to the $\frac{1}{\log n}$ power. (Note, you have to assume that $n\neq 1$ so that $\log n\neq 0$.)

Answer 2

the nub of the question is the identity (for $a,b \in \mathbb{R} \cap (1,\infty)$): $$ \log_a b \cdot \log_b a =1 \tag{1} $$ note now that by definition of a logarithm $$ n^{\log_n 10} = 10 $$ so, applying (1), we have $\log_n 10 = \frac1{\log_{10}n}$

hence: $$ n^{\frac1{\log_{10}n} } = 10 $$ to demonstrate (1) note that by definition $b= a^{\log_a b}$ and $a=b^{\log_b a}$ thus by substitution: $$ b = \left(b^{\log_b a} \right)^{{\log_a b}} = b^{\log_b a \cdot \log_a b} $$