Trouble solving for exponents with constants

Question

so I have this equation I need to solve for $i$, where $k$ and $u$ are arbitrary constants: $$ u^{{\frac1{k^i}}} = k $$ And these are the steps I've accomplished so far (all logs are in base 2): $$ log(u^{{\frac1{k^i}}}) = log(k) $$ $$ \frac1{k^i} \cdot log(u) = log(k) $$ $$ log(u) = log(k)\cdot k^i $$

But now I'm stuck... how do I solve for $i$ from here? I'm forgetting how I would do this. Any help and explanation of how log and exponent math would be great! (Also please correct me if what I've done so far is wrong.) I also hope my math formatting makes sense, thank you!

Answer 1

$$ u^{(1/k)^i} = k $$ taking logs as you have done works $$ \frac{1}{k^i}\log u = \log k $$ then we have $$ \frac{1}{k^i} = \frac{\log k}{\log u} $$ Taking logs again $$ -i\log k = \log \left(\frac{\log k}{\log u}\right) $$ so $$ i = \frac{1}{\log k}\log \left(\frac{\log u}{\log k}\right) $$

or take logs on your final statement $$ \log (\log u) = \log (\log k) + i \log k $$ we used $$ \log(ab) = \log a + \log b\\ \log(a^n) = n\log a $$

Answer 2

Instead of using a base 2 logarithm you could use base $k$, since it is implicitly non-negative. $$ log_k\left(u^{{1/k}^i}\right) = log_k(k) $$ $$ \frac{1}{k^i} log_k(u) = 1 $$ $$ log_k(u) = k^i $$ Apply $log_k$ one more time, and you're done. $$ log_k(log_k(u)) = i $$