Calculate $2^\sqrt{\log_220}-20^\sqrt{\log_{20}2}$

Question

Calculate $$2^\sqrt{\log_220}-20^\sqrt{\log_{20}2}$$

The most obvious thing that I was able to notice is that for the radicands we can write $$\log_220=\dfrac{1}{\log_{20}2}$$ So I let $\log_2{20}=a$. Then we will have $$2^\sqrt{a}-20^\sqrt{\frac{1}{a}}=2^\sqrt{a}\left(1-\dfrac{20^\sqrt{\frac{1}{a}}}{2^\sqrt{a}}\right)=2^\sqrt{a}\left(1-\dfrac{2^\frac{1}{\sqrt{a}}10^\frac{1}{\sqrt{a}}}{2^\sqrt{a}}=\right)=\\=2^\sqrt{a}\left(1-2^{\frac{1}{\sqrt{a}}-\sqrt{a}}10^\frac{1}{\sqrt{a}}\right)$$ Continuing the calculations will get us at the initial expression $2^\sqrt{a}-20^\sqrt{\frac{1}{a}}$. Something else: $$\log_220=\log_2\left(2^2.5\right)=2+\log_25$$ but I can't figure out a way to use that to solve the problem.

$$2^\sqrt{\log_220}-20^\sqrt{\log_{20}2}=0$$

Answer 1

Rewrite $20^{\sqrt{\log_{20}(2)}}$ so that it has $2$ as a base. We have \begin{align} 20^{\sqrt{\log_{20}(2)}} &= \left[2^{\log_{2}(20)} \right]^{\sqrt{\log_{20}(2)}} = 2^{\log_{2}(20)\sqrt{\log_{20}(2)}} \\&= 2^{{\log_{2}(20)}\Big/{\sqrt{\log_{2}(20)}}} = 2^{\sqrt{\log_2(20)}}. \end{align} That is, $$ 20^{\sqrt{1/a}} = [2^{a}]^{\sqrt{1/a}} = 2^{a/\sqrt{a}} = 2^\sqrt{a}. $$

Answer 2

Recall that $a^b = e^{b\ln a}$ and $\log_a b = \frac{\ln b}{\ln a}$. Thus,

$$2^\sqrt{\log_220}-20^\sqrt{\log_{20}2}$$ $$=e^{\sqrt{\log_220} \ln 2} - e^{\sqrt{\log_{20}2} \ln 20}$$ $$=e^{\sqrt{\frac{\ln 20}{\ln 2}} \ln 2} - e^{\sqrt{\frac{\ln 2}{\ln 20}} \ln 20}$$ $$=e^{\sqrt{\ln 20} \sqrt{\ln 2}} - e^{\sqrt{\ln 2}\sqrt{\ln 20}}$$ $$= 0$$

Answer 3

It seems to me that, we don't need numbers. We need prove that:

$$a^{\sqrt {\log_ab}}=b^{\sqrt {\log_ba}}$$ where $0<a,b<1\vee a,b>1$.

Let $m>0$ such that,

$$ \begin{aligned}\log_ab=m^2 \implies \begin {cases} b=a^{m^2} \\\log_ba =\frac 1{m^2}\end{cases}\end{aligned} $$

You have:

$$a^m=b^{\frac 1m}\iff a^m =a^m.$$

That's all the proof.

Answer 4

Let's think backwards.

We know that $2^{\sqrt {\log_2 20}}-20^{\sqrt {\log_{20} 2}}=0.$

So, it's enough to show that $2^{\sqrt {\log_2 20}}=20^{\sqrt{\log_{20} 2}}$.

As you said, $\log_2 20 = \dfrac 1 {\log_{20} 2}$.

So, from $2=20^{\log_{20} 2}$, $2^{\sqrt{\log_2 20}}=20^{\sqrt{log_{20} 2 \cdot \log_2 20}\cdot \sqrt{\log_{20} 2}}=20^{\sqrt{\log_{20} 2.}} \blacksquare$