Changing the base of an exponential function

Question

I don't follow the last step mentioned in the answer to this post

Why is does this hold? $$n^{k^2\log n}=2^{k^2\log^2n}$$

I would have posted this question as a comment on the post, but apparently I don't have enough reputation points to do that...

Answer 1

taking the logarithm of both sides we get $$k^2\log(n)\log(n)=k^2(\log(n))^2\log(2)$$ for $k\ne 0$ and $\log(n)\neq 0$ we get $$\frac{\log(n)}{\log(2)}=\log(n)$$ we must specify that $\log(n)=\log_2(n)$ then we get $$\log_2(n)=\log_2(n)$$

Answer 2

I think I understand it now. To get from the left side of the equation to the right side, the following substitution is being done: $$n = 2^{\log_{2} n}$$ And all of the logarithms are base 2. So then: $$n^{k^2\log_{2} n}=(2^{\log_{2} n})^{k^2\log_{2} n}$$ which becomes: $$2^{(\log_{2} n) k^2\log_{2} n}$$ which is: $$2^{k^2(\log_{2} n)^2}$$ which is the same as: $$2^{k^2\log^2 n}$$ (where the base of the logarithm is 2).

Thanks for all of the replies!

Answer 3

It should be $$n^{k^2\ln{n}}=\left(2^{\log_2n}\right)^{k^2\ln{n}}=\left(2^{\frac{\ln{n}}{\ln2}}\right)^{k^2\ln{n}}=2^{\frac{1}{\ln2}k^2\ln^2n}$$

Answer 4

I guess I am not wrong here, if I take 2 as the base of the natural logarithm:$$2^{{log_2 n}^{k^2 log n}}=2^{k^2(logn) (log n)}=2^{(logn)^2 k^2}$$