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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?

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2 Answers 2

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We have:

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

Hence:

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

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    $\begingroup$ @Bachelier : 2 things to note re player3236's answer (1) his formula is distinct from $\log A - \log B = \log \left(\frac{A}{B}\right)$. (2) His formula can be proven by noticing that $a^{[(\log_a c) \times (\log_c b)]} = [a^{(\log_a c)}]^ {(\log_c b)} = c^{(\log_c b)} = b = a^{(\log_a b)}.$ Therefore, $(\log_a c) \times (\log_c b) = (\log_a b).$ $\endgroup$ Oct 1, 2020 at 11:41
  • $\begingroup$ or $\log_a b = \log_a \left(c^{\textstyle\log_c b}\right) = \log_c b \log_a c$ $\endgroup$
    – player3236
    Oct 1, 2020 at 11:50
  • $\begingroup$ elegant alternative $\endgroup$ Oct 1, 2020 at 13:23
  • $\begingroup$ That was it. Change of base formula. Thank you. $\endgroup$
    – Bachelier
    Oct 1, 2020 at 13:31
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As an alternative, using $A^B=e^{B\ln A}$, we have

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

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