QJEEM23Ma1541: JEE Main 2023 Mathematics - Class 12 JEEM23Ma1541 Solution: Let a, b, c be three distinct positive real numbers such that…

Given, $(2 a)^{\ln a}=(b c)^{\ln b}$ , where 2 a>0, b c>0

$\Rightarrow \ln a(\ln 2+\ln a)=\ln b(\ln b+\ln c)$ ..........(i)

and $(b)^{\ln 2}=(a)^{\ln c}$

$\Rightarrow \ln 2 \cdot \ln b=\ln c \cdot \ln a$ ..........(ii)

Now, let $\ln a=x, \ln b=y$

$\ln 2=p, \ln c=z$

Now, from Eqs. (i) and (ii), we get

$p \cdot y=x z \text { and } x(p+x)=y(y+z)$

$\begin{aligned}& \therefore p=\frac{x z}{y} \\\\& \therefore x\left(\frac{x z}{y}+x\right)=y(y+z) \\\\& \Rightarrow x^2 z+x^2 y=y^2(y+z) \\\\& \Rightarrow \left(x^2-y^2\right)(y+z)=0\end{aligned}$

$\therefore$ $x^2 = y^2$

$x = \pm y$

So, from the equations, there are two cases :

Case 1 :

$x = y$

In this case, since x and y are natural logarithms of positive numbers a and b respectively, this implies that a = b. However, this cannot be true as a, b, and c are given to be distinct positive real numbers.

Case 2 :

x = -y

In this case, $\ln a = -\ln b$

$\Rightarrow a \times b = 1$

$\Rightarrow b = \frac{1}{a}$

Also, y + z = 0 (from the equations above)

$\Rightarrow \ln b + \ln c = 0$

$\Rightarrow \ln (b \times c) = 0$

$\Rightarrow b \times c = 1$

Given $b = \frac{1}{a}$ and $b \times c = 1$ $\Rightarrow c = a$

Thus, in the case where x = -y, the possible values are :

$b = \frac{1}{a}$

$c = a$

$\begin{aligned}& \text { If } b c=1 \Rightarrow(2 a)^{\ln a}=1 \\\\& \Rightarrow a=1 / 2 \\\\& \text { So, } 6 a+5 b c=6\left(\frac{1}{2}\right)+5=3+5=8\end{aligned}$