0
$\begingroup$

I have the expression $2^{\sqrt{\log(n)}}$, but that's nasty to work with. I watched a few video on logarithms https://www.youtube.com/watch?v=ZIwmZ9m0byI , but none of it seems to cover how I might break the log out?
What are the steps to breaking this apart into something a little more workable?

$\endgroup$
  • $\begingroup$ Please in the future use MathJax to type formulas $\endgroup$ – Yuriy S Jan 23 '18 at 23:12
  • $\begingroup$ As for the question, you can't 'break the logarithm out' because of the square root. By the way, what do you need this expression for? What do you mean by "work with"? $\endgroup$ – Yuriy S Jan 23 '18 at 23:13
  • $\begingroup$ I attempted to write this in mathjax, however it would not format correctly. I'll save your edit for future reference $\endgroup$ – Podo Jan 24 '18 at 0:13
  • $\begingroup$ @Podo Please, if you are ok, you can accept the answer and set it as solved. Thanks! cdn.sstatic.net/img/faq/faq-accept-answer.png $\endgroup$ – gimusi Feb 8 '18 at 22:05
1
$\begingroup$

I think that there are not algebraic manipulation to simplify it.

You can change the base as follow

$$2^{\sqrt{\log n}}=e^{\sqrt{\log n}\cdot \log 2}$$

$\endgroup$
1
$\begingroup$

Perhaps it would be helpful if you set the expression equal to another variable. I've assumed you're using log of base 10, but it will also be similar in another base.

Given: $y = 2^{\sqrt{log_{10}(n)}}$.

This can be rewritten as:

$log_2(y) = \sqrt{log_{10}(n)}$

$(log_2(y))^2 = log_{10}(n)$

$10^{({log_2(y)})^2} = n$

$\endgroup$
  • $\begingroup$ I'm using log2 (comparing complexity of algorithmic run times) but this is, i think at least, what I was looking for. $\endgroup$ – Podo Jan 24 '18 at 0:15
  • $\begingroup$ Sounds good. That would change it to be: $2^{(log_2(y))^2}=n.$ Which I think would simplify to: $\endgroup$ – E. Tucker Jan 24 '18 at 12:48
  • $\begingroup$ Sounds good. That would change it to be: $2^{(log_2(y))^2}=n.$ Which I think would simplify to: $2^{log_2(y)*log_2(y)} = (2^{log_2y})^{log_2(y)}=y^{log_2(y)}$. $\endgroup$ – E. Tucker Jan 24 '18 at 12:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.