Question

Let a, b, c be three distinct positive real numbers such that (2a)logea = (bc)logeb and bloge2 = alogec. Then 6a + 5bc is equal to _____.

Answer 1

Given :
(2a)ln a = (bc)ln b 2a > 0, bc > 0 bln2 = alnc
ln a(ln2 + ln a) = ln b (ln b + ln c) ln2.lnb = lnc.lna
ln 2 = a, ln a = x1 ln b = y, ln c = z ay = yz
x(a + x) = y(y + 2)
$a=\frac{xy}{y}$ $(2a)^{\text{lna}}=(2a)^0$
$x(\frac{xy}{y}+x)=y(y+z)$
x2(z + y) = y2(y + z)
y + z = 0 or x2 = y2 ⇒ x = -y
bc = 1 or ab = 1

$(a,b,c)=(\frac{1}{2},\lambda,\frac{1}{\lambda}),\lambda\ne1,2,\frac{1}{2}$
then 6a + 5bc = 3 + 5 = 8
So, the correct answer is 8.