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Functions — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsFunctions5 Marks
Question
Solve: $\sqrt{\log _2 x^4}+4 \log _4 \sqrt{\frac{2}{x}}=2$
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Answer

$ \sqrt{\log _2 x^4}+4 \log _4 \sqrt{\frac{2}{x}}=2$
$\therefore \quad \sqrt{\log _2 x^4}+4 \log _4\left(\frac{2}{x}\right)^{\frac{1}{2}}=2$
$\therefore \quad \sqrt{4 \log _2 x}+\frac{4}{2} \log _4\left(\frac{2}{x}\right)=2$
$\therefore \quad 2 \sqrt{\log _2 x}+2 \log _4\left(\frac{2}{x}\right)=2$
$\therefore \quad \sqrt{\log _2 x}+\log _4\left(\frac{2}{x}\right)=1$
$\therefore \quad \sqrt{\log _2 x}+\frac{\log _2\left(\frac{2}{x}\right)}{\log _2 4}=1$
$\therefore \quad \sqrt{\log _2 x}+\frac{\log _2\left(\frac{2}{x}\right)}{\log _2(2)^2}=1$
$\therefore \quad \sqrt{\log _2 x}+\frac{\log _2 2-\log _2 x}{2 \log _2 2}=1$
$\therefore \quad \sqrt{\log _2 x}+\frac{1-\log _2 x}{2(1)}=1 \quad \ldots\left[\because \log _a a=1\right]$
Let $\log _2 x=\mathrm{a}$
$\therefore \quad \sqrt{\mathrm{a}}+\frac{1-\mathrm{a}}{2}=1$
Multiplying throughout by 2 , we get
$ 2 \sqrt{a}+1-a=2$
$\therefore 2 \sqrt{a}=a+1$
Squaring on both sides, we get
$ 4 a=(a+1)^2$
$\therefore 4 a=a^2+2 a+1$
$\therefore a^2-2 a+1=0$
$\therefore (a-1)^2=0$
$\therefore a-1=0$
$\therefore a=1 $
Since $\log _2 x=\mathrm{a}$,
$\log _2 x=1$
$\therefore \quad x=2^1$
$\therefore \quad x=2$

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