Let a,b,c be the three distinct positive real numbers such that (2a)^ log_ea=(bc)^ log_eb and b^ log_e2=a^ log_ec Then 6a+5bc is equal to .

Given equation: ${(2a)}^{\mathrm{ln}(a)}={(bc)}^{\mathrm{ln}(b)}$(2a)ln(a)=(bc)ln(b)
$\Rightarrow \mathrm{ln}(a)\cdot (\mathrm{ln}(2a))=\mathrm{ln}(b)\cdot \mathrm{ln}(bc)$⇒ln(a)⋅(ln(2a))=ln(b)⋅ln(bc)
$\Rightarrow \mathrm{ln}(a)\cdot (\mathrm{ln}(2)+\mathrm{ln}(a))=\mathrm{ln}(b)\cdot (\mathrm{ln}(b)+\mathrm{ln}(c))$⇒ln(a)⋅(ln(2)+ln(a))=ln(b)⋅(ln(b)+ln(c))
let $\mathrm{ln}(a)=x,\mathrm{ln}(b)=y,\mathrm{ln}(c)=z,x\neq y\neq z$ln(a)=x,ln(b)=y,ln(c)=z,x=y=z
$\Rightarrow x(\mathrm{ln}(2)+x)=y(y+z)$⇒x(ln(2)+x)=y(y+z)
$\Rightarrow x\mathrm{ln}(2)={y}^{2}-{x}^{2}+yz$⇒xln(2)=y2−x2+yz ........(i)
Similarly, from second equation
$\mathrm{ln}(2)\cdot \mathrm{ln}(b)=\mathrm{ln}(c)\cdot \mathrm{ln}(a)$ln(2)⋅ln(b)=ln(c)⋅ln(a)
$\Rightarrow \mathrm{ln}(2)=\frac{xz}{y}$⇒ln(2)=yxz​ ......(ii)
Substitute $eq(ii)\text{in}eq(i)$eq(ii)ineq(i),
We get,
$\Rightarrow {y}^{3}-y{x}^{2}+{y}^{2}z={x}^{2}z$⇒y3−yx2+y2z=x2z
$\Rightarrow {y}^{2}(y+z)-{x}^{2}(y+z)=0$⇒y2(y+z)−x2(y+z)=0
$\Rightarrow ({y}^{2}-{x}^{2})(y+z)=0$⇒(y2−x2)(y+z)=0
$\Rightarrow (x-y)(x+y)(y+z)=0$⇒(x−y)(x+y)(y+z)=0
$\because x\neq y\Rightarrow (x+y)(y+z)=0$∵x=y⇒(x+y)(y+z)=0
Now $x=-y,y=-z\text{and}x=z$x=−y,y=−zandx=z
But $\mathrm{ln}(2)=\frac{xz}{y}$ln(2)=yxz​
$\Rightarrow \mathrm{ln}(2)=\frac{x(x)}{-x}$⇒ln(2)=−xx(x)​
$\Rightarrow \mathrm{ln}(2)=-\mathrm{ln}(a)$⇒ln(2)=−ln(a)
$=-\mathrm{ln}2\Rightarrow a=\frac{1}{2}$=−ln2⇒a=21​ and
$\Rightarrow y=-z$⇒y=−z
$\Rightarrow \mathrm{ln}(b)=-\mathrm{ln}(c)$⇒ln(b)=−ln(c)
$\Rightarrow bc=1$⇒bc=1
Now $6a+5bc=6(\frac{1}{2})+5(1)=8$6a+5bc=6(21​)+5(1)=8
Hence this is the required answer.