# Trouble solving for exponents with constants

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!

• i don't understand what exactly do you mean? – Dr. Sonnhard Graubner Feb 27 '17 at 19:59
• instead of having the first equation = k, I would like it to = i. In otherwords, I'm trying to solve for i. – keenns12 Feb 27 '17 at 20:06
• $\log u = \log k*k^i$ so $k^i = \frac {\log u}{\log k}$ so $i = \log_k(\frac {\log u}{\log k})$. That's $i$. That's all there is. You are allowed to express logs of logs, you know. – fleablood Feb 27 '17 at 20:13

$$u^{(1/k)^i} = k$$ 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 $$\log(ab) = \log a + \log b\\ \log(a^n) = n\log a$$
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. $$log_k(log_k(u)) = i$$