Question

If $\alpha ,\beta$ are the roots of equation ${\log }_5(5^{\frac{1}{x}}+125)={\log }_{5}6+1+\frac{1}{2x}$ then evaluate the sum $(\frac{1}{\alpha }+\frac{1}{\beta })$

Answer 1

${\log }_5(5^{\frac{1}{x}}+125)={\log }_{5}6+1+\frac{1}{2x}$

${\log }_5(5^{\frac{1}{x}}+125)={\log }_{5}6+{\log }_{5}5+\frac{1}{2x}$

${\log }_5(5^{\frac{1}{x}}+125)={\log }_{5}30+\frac{1}{2x}$

${\log }_5(5^{\frac{1}{x}}+125)-{\log }_{5}30=\frac{1}{2x}$

${\log }_5\left(\frac{5^{\frac{1}{x}}+125}{30}\right)=\frac{1}{2x}$

$\frac{5^{\frac{1}{x}}+125}{30}=5^{\frac{1}{2x}}$

$5^{\frac{1}{x}}+125=30\times 5^{\frac{1}{2x}}$

$Let$ $5^{\frac{1}{2x}}=k$ then $5^{\frac{1}{x}}=k^2$

$\therefore k^2+125=30k$

$\Rightarrow k^2-30k+125=0$

$\Rightarrow k^2-25k-5k+125=0$

$\Rightarrow k(k-25)-5(k-25)=0$

$\Rightarrow (k-25)(k-5)=0$

$\Rightarrow k=25\left(or\right)k=5$

i.e., $5^{\frac{1}{2x}}=25=5^2\left(or\right)$ $5^{\frac{1}{2x}}=5=5^1$

$\frac{1}{2x}=2$ (or) $\frac{1}{2x}=1$

$x=\frac{1}{4}$ (or) $x=\frac{1}{2}$

i.e., $\alpha =\frac{1}{4}$ and $\beta =\frac{1}{2}$

$\frac{1}{\alpha }=4$ and $\frac{1}{\beta }=2$

$\therefore (\frac{1}{\alpha }+\frac{1}{\beta })$ $=4+2=6$.