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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!

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  • $\begingroup$ i don't understand what exactly do you mean? $\endgroup$ – Dr. Sonnhard Graubner Feb 27 '17 at 19:59
  • $\begingroup$ instead of having the first equation = k, I would like it to = i. In otherwords, I'm trying to solve for i. $\endgroup$ – keenns12 Feb 27 '17 at 20:06
  • $\begingroup$ $\log u = \log k*k^i$ 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. $\endgroup$ – fleablood Feb 27 '17 at 20:13
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$$ u^{(1/k)^i} = k $$ 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 $$ \log(ab) = \log a + \log b\\ \log(a^n) = n\log a $$

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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. $$ log_k(log_k(u)) = i $$

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