JEE Main 2023 — Logarithm Question with Solution

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

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

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

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

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

$x=\pm y$

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 :

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

$c=a$

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