Question

Given that $a$ and $b$ are positive numbers satisfying the equation $4({\log }_{10}a{)}^2+({\log }_{2}b{)}^2=1$ , then show that $a\in [\frac{1}{\sqrt{10}},\sqrt{10}],b\in [\frac{1}{10},10]$.

Answer 1

Given, $4({\log }_{10}a{)}^2+({\log }_{2}b{)}^2=1$

$\Rightarrow (2{\log }_{10}a{)}^2+({\log }_{2}b{)}^2=1$

We know that ${\cos }^2\theta +{\sin }^2\theta =1$

Comparing this with given equation,

$2{\log }_{10}a=\cos \theta ;{\log }_{2}b=\sin \theta$

${\log }_{10}a=\frac{\cos \theta }{2}$

Since, $-1\le \cos \theta \le 1$

$\Rightarrow -1\le 2{\log }_{10}a\le 1$
$\Rightarrow \frac{-1}{2}\le {\log }_{10}a\le \frac{1}{2}$

$\Rightarrow {10}^{-1/2}\le a\le {10}^{1/2}$
$\Rightarrow a\in [\frac{1}{\sqrt{10}},\sqrt{10}]$

Also since, $-1\le \sin \theta \le 1$

$\Rightarrow -1\le {\log }_{10}b\le 1$

$\Rightarrow {10}^{-1}\le b\le 10$
$\Rightarrow b\in [\frac{1}{10},10]$