Question

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

• A. $\frac{1}{x_0}+\frac{1}{y_0}=7$
• B. $\frac{1}{x_0}-\frac{1}{y_0}=1$
• C. $4x_0-3y_0=1$
• D. $x_0+y_0=\frac{7}{12}$

Answer 1

Answer: (A)

$3^{\ln x}=4^{\ln y}$ .....(1)
and $(4x{)}^{\ln 4}=(3y{)}^{\ln 3}$ .....(2)
$(1)\Rightarrow (\ln x)(\ln 3)=(\ln y)(\ln 4)$
$(2)\Rightarrow (\ln 4)(\ln 4+\ln x)=(\ln 3)(\ln 3+\ln y)$
${\ln }^{2}4-{\ln }^{2}3=(\ln 3)\ln y-(\ln 4)\ln x$
${\ln }^{2}4-{\ln }^{2}3=\frac{\left({\ln }^{2}3\right)(\ln x)-{\ln }^{2}4(\ln x)}{\ln 4}$
$\therefore \ln x=-\ln 4$
$\Rightarrow x_0=\frac{1}{4}\text{ and }y=\frac{1}{3}.$