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Functions — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsFunctions3 Marks
Question
Solve for $\mathrm{x}$ :$2 \log _{10} x=1+\log _{10}\left(x+\frac{11}{10}\right)$
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Answer

$ 2 \log _{10} x=1+\log _{10}\left(x+\frac{11}{10}\right)$
$\log _{10} x^2-\log _{10}\left(x+\frac{11}{10}\right)=1$
$\ldots\left[\text { log } \mathrm{m}=\log \mathrm{m}^{\mathrm{n}}\right]$
$\therefore \quad \log _{10} x^2-\log _{10}\left(\frac{10 x+11}{10}\right)=1$
$\therefore \quad \log _{10}\left(\frac{x^2}{\frac{10 x+11}{10}}\right)_{}=1 \quad \ldots\left[\log \mathrm{m}-\log \mathrm{n}=\log \frac{\mathrm{m}}{\mathrm{n}}\right]$
$\therefore \quad \log _{10}\left(\frac{10 x^2}{10 x+11}\right)=\log _{10} 10 \quad \ldots\left[\log _{\mathrm{a}} \mathrm{a}=1\right]$
$\therefore \quad \frac{10 x^2}{10 x+11}=10$
$\therefore \quad \frac{x^2}{10 x+11}=1$
$\therefore x^2=10 \mathrm{x}+11$
$\therefore x^2-10 x-11=0$
$\therefore(x-11)(\mathrm{x}+1)=0$
$\therefore x=11 \text { or } \mathrm{x}=-1 $
But $\log$ of a negative numbers does not exist
$ \therefore \mathrm{x} \neq-1$
$\therefore \mathrm{x}=11 $

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