Exponent Simplification

Question

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!

Answer 1

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 $u^{\log_u(k)}=k$,

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

Answer 2

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

Answer 3

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.