Let a, b, c be three distinct positive real numbers such that [(2a)^ _ea = (bc)^ _eb] and …

Question

Let a, b, c be three distinct positive real numbers such that $${(2a)^{{{\log }_e}a}} = {(bc)^{{{\log }_e}b}}$$ and $${b^{{{\log }_e}2}} = {a^{{{\log }_e}c}}$$.Then, 6a + 5bc is equal to \_\_\_\_\_\_\_\_\_\_\_.

Answer: 8

Accepted Answer

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

$p\cdot y=xzx(p+x)=y(y+z)$

$\begin{gathered} 	herefore p=\frac{xz}{y} \ 	herefore x(\frac{xz}{y}+x)=y(y+z) \ \Rightarrow {x}^{2}z+{x}^{2}y={y}^{2}(y+z) \ \Rightarrow ({x}^{2}-{y}^{2})(y+z)=0 \end{gathered}$
$	herefore $ ${x}^{2}={y}^{2}$

$x=\pm y$

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, $\ln a=-\ln b$

$\Rightarrow a	imes b=1$

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

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

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

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

$\Rightarrow b	imes c=1$

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

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

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

$c=a$

$\begin{gathered} bc=1\Rightarrow (2a{)}^{\ln a}=1 \ \Rightarrow a=1/2 \ 6a+5bc=6(\frac{1}{2})+5=3+5=8 \end{gathered}$

Solutions · 1