Let a, b, c be three distinct positive real numbers such that ( 2 a ) log e a = ( b c ) log e b and b log e 2 = a log e c . Then 6a + 5bc is equal to _____________ . [2023]

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Solution

Q.
Let a, b, c be three distinct positive real numbers such that ${(2 a)}^{\log_{e} a} = {(b c)}^{\log_{e} b}$ and $b^{\log_{e} 2} = a^{\log_{e} c}$ .

Then 6a + 5bc is equal to _____________ . [2023]

Ans.
(8)

${(2 a)}^{\log_{e} a} = {(b c)}^{\log_{e} b}, where 2 a > 0 and b c > 0$

$\Rightarrow \log_{e} a (\log_{e} 2 + \log_{e} a) = \log_{e} b (\log_{e} b + \log_{e} c)$ ...(i)

Also, $b^{\log_{e} 2} = a^{\log_{e} c}$

$\Rightarrow \log_{e} 2 \log_{e} b = \log_{e} c \log_{e} a$ ...(ii)

Let $\log_{e} 2 = α, \log_{e} a = x, \log_{e} b = y, and \log_{e} c = z$

$∴ From (i) and (ii), we get x (α + x) = y (y + z) and α y = x z$

$\Rightarrow x (\frac{x z}{y} + x) = y (y + z)$ $[∵ α = \frac{x z}{y}]$

$\Rightarrow x^{2} (z + y) = y^{2} (y + z)$

$\Rightarrow y + z = 0 or x^{2} = y^{2} \Rightarrow x = - y$ $[∵ x = y is not possible]$

$\Rightarrow b c = 1 or a b = 1$

Now, $b c = 1 \Rightarrow 2 a \log_{e} a = 1 \Rightarrow a = 1 or \frac{1}{2}$

$Case - 1: Distinct possible values of a, b, c can be (\frac{1}{2}, p, \frac{1}{p}), where p \neq 1, 2, \frac{1}{2}$

$∴ 6 a + 5 b c = 6 (\frac{1}{2}) + 5 (p) (\frac{1}{p}) = 3 + 5 = 8$

$Case - 2: (a, b, c) can be (p, \frac{1}{p}, \frac{1}{2}), where p \neq (1, 2, \frac{1}{2})$

So, in this case infinite answers are possible.

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