Exponents and Log properties

Question

This is part of a proof I am working on.

So on one line, I needed to show that $$2^{\Bigg(\cfrac{\ln\frac{1}{n}}{\ln(2)}\Bigg)} =\frac{1}{n}$$

That was $2$ to the power of $\ \ \cfrac{\ln\frac{1}{n}}{\ln(2)}$

Now I tried to use the log/exponent cancellation techniques to no avail. I think there is a property I am missing.

Any suggestions?

Answer 1

We have:

$$\frac{\log_a b}{\log_a c} = \log_c b$$

Hence:

$$2^{\Bigg(\cfrac{\ln\frac{1}{n}}{\ln(2)}\Bigg)}=2^{\textstyle\log_2\frac{1}{n}} =\frac{1}{n}$$

Answer 2

As an alternative, using $A^B=e^{B\ln A}$, we have

$$\large 2^{\left(\frac{\ln\frac{1}{n}}{\ln 2}\right)}=e^{\left(\frac{\ln\frac{1}{n}}{\ln 2}\ln 2\right)}=e^{\ln \frac1n}=\frac1n$$