Let a, b, c be three distinct positive real n | vedclass

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(A) Given equations are $(2a)^{\ln a} = (bc)^{\ln b}$ and $b^{\ln 2} = a^{\ln c}$.Taking the natural logarithm on both sides of the first equation: $\

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(A) Given equations are $(2a)^{\ln a} = (bc)^{\ln b}$ and $b^{\ln 2} = a^{\ln c}$.Taking the natural logarithm on both sides of the first equation: $\ln a (\ln 2 + \ln a) = \ln b (\ln b + \ln c)$.From the second equation: $\ln 2 \cdot \ln b = \ln c \cdot \ln a \implies \ln c = \frac{\ln 2 \cdot \ln b}{\ln a}$.Substituting $\ln c$ into the first equation: $(\ln a)^2 + \ln a \ln 2 = (\ln b)^2 + \ln b \left( \frac{\ln 2 \cdot \ln b}{\ln a} ight)$.$(\ln a)^2 + \ln a \ln 2 = (\ln b)^2 \left( \frac{\ln a + \ln 2}{\ln a} ight)$.$(\ln a)^2 (\ln a + \ln 2) = (\ln b)^2 (\ln a + \ln 2)$.Since $a, b, c$ are distinct,$\ln a + \ln 2 
eq 0$,so $(\ln a)^2 = (\ln b)^2$,which implies $\ln a = -\ln b$ (as $a 
eq b$).Thus $b = 1/a$. Substituting into the second equation: $(1/a)^{\ln 2} = a^{\ln c} \implies a^{-\ln 2} = a^{\ln c} \implies \ln c = -\ln 2 \implies c = 1/2$.Substituting $b = 1/a$ and $c = 1/2$ into the first equation: $(2a)^{\ln a} = (a/2)^{\ln(1/a)} = (a/2)^{-\ln a} = (2/a)^{\ln a}$.Since $(2a)^{\ln a} = (2/a)^{\ln a}$,we have $2a = 2/a \implies a^2 = 1$. Since $a > 0$,$a = 1$. If $a=1$,then $b=1$,which contradicts that $a, b, c$ are distinct. The problem statement is mathematically inconsistent.

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

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