Question
$\log _2 x+\log _4 x+\log _{16} x=\frac{21}{4}$
✓
Answer
$\log _2 x+\log _4 x+\log _{16} x=\frac{21}{4} $
$ \therefore \frac{1}{\log _x 2}+\frac{1}{\log _x 2^2}+\frac{1}{\log _x 2^4}=\frac{21}{4} $
$ \therefore \frac{1}{\log _x 2}+\frac{1}{2 \log _x 2}+\frac{1}{4 \log _x 2}=\frac{21}{4} $
$ \therefore \frac{1}{\log _x 2}\left(1+\frac{1}{2}+\frac{1}{4}\right)=\frac{21}{4} $
$ \therefore \frac{1}{\log _x 2}\left(\frac{7}{4}\right)=\frac{21}{4} $
$\therefore \log _x 2=\frac{7}{4} \cdot \frac{4}{21}$
$ \therefore \log _x 2=\frac{1}{3}$
$\therefore x^{\frac{1}{3}}=2$
$ \therefore x=2^3 $
$ =8 .$
$ \therefore \frac{1}{\log _x 2}+\frac{1}{\log _x 2^2}+\frac{1}{\log _x 2^4}=\frac{21}{4} $
$ \therefore \frac{1}{\log _x 2}+\frac{1}{2 \log _x 2}+\frac{1}{4 \log _x 2}=\frac{21}{4} $
$ \therefore \frac{1}{\log _x 2}\left(1+\frac{1}{2}+\frac{1}{4}\right)=\frac{21}{4} $
$ \therefore \frac{1}{\log _x 2}\left(\frac{7}{4}\right)=\frac{21}{4} $
$\therefore \log _x 2=\frac{7}{4} \cdot \frac{4}{21}$
$ \therefore \log _x 2=\frac{1}{3}$
$\therefore x^{\frac{1}{3}}=2$
$ \therefore x=2^3 $
$ =8 .$
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