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I have this expression where u and k are arbitrary constants: $$ \large u^{\left({\frac1{k}}\right)^{\log_k(\log_k(u))}} $$

I'm trying to clean up or simplify this expression... how can i go just making this cleaner? I'm forgetting log and exponent rules. Thanks!

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    Try "\log_k x" to write $\log_k x$ – 2017-02-27
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    I enlarged the formula to make it visible that there are two layers of exponents here. Is that what you wanted? I.e, is the exponent of $u$, meant to be a power: $\left(\frac{1}{k}\right)^{\log_k \ldots}$ or a product: $\left(\frac{1}{k}\right)\log_k \ldots$? – 2017-02-27
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    Using technology I got $\large u^{\left({\frac1{k}}\right)^{\log_k(\log_k(u))}} = u^{{\frac1{\log_k(u)}}}$ – 2017-02-27
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    @WaveX Hm, it seems technology isn't very good at algebra. :-( – 2017-02-27
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    @SimplyBeautifulArt I checked it using both my TI-89 and Wolfram Alpha, using Wolfram Alpha it simplifies to $k$, however the TI-89 produces the solution I gave above. It seem that it is still equal, just didn't finish simplifying – 2017-02-27
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    XD I wasn't saying it was wrong, I would just expect it to simplify more. @WaveX – 2017-02-27

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Note that

$$\left(\frac1k\right)^{\log_k(\log_k(u))}=\frac1{k^{\log_k(\log_k(u))}}$$

Since $k^{\log_k(a)}=a$, then

$$\frac1{k^{\log_k(\log_k(u))}}=\frac1{\log_k(u)}$$

Since $1=\log_k(k)$ and by change of base formula $\frac{\log_b(p)}{\log_b(q)}=\log_q(p)$, then

$$\frac1{\log_k(u)}=\frac{\log_k(k)}{\log_k(u)}=\log_u(k)$$

Recalling once again that $u^{\log_u(k)}=k$,

$$u^{\left(\frac1k\right)^{\log_k(\log_k(u))}}=u^{\log_u(k)}=k$$

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$$(\frac{1}{k})^{\log_{k}(\log_{k}(u))}=k^{-\log_{k}(\log_{k}(u))}=$$ $$=k^{\log_{k}(\frac{1}{\log_{k}(u)})}=\frac{1}{\log_{k}(u)}=\log_{u}(k)$$ then $$u^{(\frac{1}{k})^{\log_{k}(\log_{k}(u))}}=k$$

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Let $k^x=u$ and let $k^y=x$. Then, the expression simplifies to $u^{(\frac{1}{k})^{\log_k x}}=u^{(\frac{1}{k})^y}=u^{\frac{1}{k^y}}$. However, we know that $k^y=x$, so $u^{\frac{1}{x}}$.

However, note that since $k^x=u$, $k=u^{\frac{1}{x}}$. Therefore, the entire expression simplifies down to just $k$.

Note that this solution did not use any super complicated laws of exponents. In general, you can feel your way to a solution using substitutions to get rid of those pesky logs and exponents.