Question

Solve $3{\log }_{x}4+2{\log }_{4x}4+3{\log }_{16x}4=0$ is $x=\frac{1}{a}\text{and}x=\frac{1}{a^3}$ Find $a$.

Answer 1

$3{\log }_{x}4+2{\log }_{4x}4+3{\log }_{16x}4=0$

$\Rightarrow \frac{3}{{\log }_{4}x}+\frac{2}{{\log }_{4}4x}+\frac{3}{{\log }_{4}16x}=0$ (as ${\log }_{a}b=\frac{1}{{\log }_{b}a}$)

$\Rightarrow \frac{3}{{\log }_{4}x}+\frac{2}{{\log }_{4}4+{\log }_{4}x}+\frac{3}{{\log }_{4}16+{\log }_{4}x}=0$ (as $\log ab=\log a+\log b)$

let ${\log }_{4}x=y$

$\Rightarrow \frac{3}{y}+\frac{2}{1+y}+\frac{3}{2+y}=0$ ($\log a^n=n\log a\&{\log }_{a}a=1$)

$\Rightarrow 3\left(y+1\right)\left(y+2\right)+2y\left(y+2\right)+3y\left(y+1\right)=0$

$\Rightarrow 3\left(y^2+3y+2\right)+2y^2+4y+3y^2+3y=0$

$\Rightarrow 4y^2+8y+3=0$

$\Rightarrow 4y^2+6y+2y+3=0$

$\Rightarrow 2y\left(2y+3\right)+1\left(2y+3\right)=0$

$\Rightarrow \left(2y+3\right)\left(2y+1\right)=0$

$\Rightarrow y=\frac{-3}{2}$ or $y=\frac{-1}{2}$

If $y=\frac{-3}{2}$

$\Rightarrow {\log }_{4}x=\frac{-3}{2}$

$\Rightarrow x=4^{\frac{-3}{2}}$

$\Rightarrow x=\frac{1}{8}$

$\Rightarrow x=\frac{1}{2^3}$

and if $y=\frac{-1}{2}$

$\Rightarrow {\log }_{4}x=\frac{-1}{2}$

$\Rightarrow x=4^{\frac{-1}{2}}$

$\Rightarrow x=\frac{1}{2}$

on comparing with $x=\frac{1}{a}$ or $x=\frac{1}{a^3}$

we get $a=2$