Let a, b, c be real numbers, each greater tha | vedclass

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(D) Given the equation: $\frac{2}{3} \log _{b} a+\frac{3}{5} \log _{c} b+\frac{5}{2} \log _{a} c=3$.Let $x = \frac{2}{3} \log _{b} a$, $y = \frac{3}{5

Vedclass · Accepted Answer

$27$

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(D) Given the equation: $\frac{2}{3} \log _{b} a+\frac{3}{5} \log _{c} b+\frac{5}{2} \log _{a} c=3$.Let $x = \frac{2}{3} \log _{b} a$, $y = \frac{3}{5} \log _{c} b$, and $z = \frac{5}{2} \log _{a} c$.Note that $x \cdot y \cdot z = (\frac{2}{3} \log _{b} a) \cdot (\frac{3}{5} \log _{c} b) \cdot (\frac{5}{2} \log _{a} c) = (\frac{2}{3} \cdot \frac{3}{5} \cdot \frac{5}{2}) \cdot (\log _{b} a \cdot \log _{c} b \cdot \log _{a} c) = 1 \cdot 1 = 1$.By the Arithmetic Mean-Geometric Mean inequality $(AM \ge GM)$, for positive real numbers $x, y, z$ with $x+y+z=3$ and $xyz=1$, the equality holds only when $x=y=z=1$.Thus, $\frac{2}{3} \log _{b} a = 1 \Rightarrow \log _{b} a = \frac{3}{2}$.Given $b=9$, we have $\log _{9} a = \frac{3}{2}$.Therefore, $a = 9^{3/2} = (3^2)^{3/2} = 3^3 = 27$.

Let $a, b, c$ be real numbers, each greater than $1$, such that $\frac{2}{3} \log _{b} a+\frac{3}{5} \log _{c} b+\frac{5}{2} \log _{a} c=3$. If the value of $b$ is $9$, then the value of $a$ must be

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