Question

If $\left(x_0, y_0\right)$ be the solution of the system of the equation $3^{\ln x}=4^{\ln y}$ and $(4 x)^{\ln 4}=(3 y)^{\ln 3}$, then

• A. $ \frac{1}{ x _0}+\frac{1}{ y _0}=7 $
• B. $ \frac{1}{ x _0}-\frac{1}{ y _0}=1 $
• C. $ 4 x _0-3 y _0=1 $
• D. $ x _0+ y _0=\frac{7}{12}$

Answer 1

Answer: (A)

$3^{\ln x}=4^{\ln y}$ .....(1)
and $(4 x )^{\ln 4}=(3 y )^{\ln 3}$ .....(2)
$(1) \Rightarrow(\ln x)(\ln 3)=(\ln y)(\ln 4)$
$(2) \Rightarrow(\ln 4)(\ln 4+\ln x)=(\ln 3)(\ln 3+\ln y)$
$\ln ^2 4-\ln ^2 3=(\ln 3) \ln y -(\ln 4) \ln x $
$\ln ^2 4-\ln ^2 3=\frac{\left(\ln ^2 3\right)(\ln x )-\ln ^2 4(\ln x )}{\ln 4} $
$\therefore \ln x =-\ln 4$
$\Rightarrow x _0=\frac{1}{4} \text { and } y =\frac{1}{3} .$