Question

Let a, b, c be 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 _____.

Answer 1

Given :
(2a)^{ln a} = (bc)^{ln b} 2a > 0, bc > 0  b^ln2 = a^lnc
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.