Let m be the minimum possible value of log _3 | vedclass

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(B) By the $AM$-$GM$ inequality,$\frac{3^{y_1}+3^{y_2}+3^{y_3}}{3} \geq \sqrt[3]{3^{y_1} \cdot 3^{y_2} \cdot 3^{y_3}} = \sqrt[3]{3^{y_1+y_2+y_3}}$.Giv

Vedclass · Accepted Answer

$8$

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(B) By the $AM$-$GM$ inequality,$\frac{3^{y_1}+3^{y_2}+3^{y_3}}{3} \geq \sqrt[3]{3^{y_1} \cdot 3^{y_2} \cdot 3^{y_3}} = \sqrt[3]{3^{y_1+y_2+y_3}}$.Given $y_1+y_2+y_3=9$,we have $\frac{3^{y_1}+3^{y_2}+3^{y_3}}{3} \geq \sqrt[3]{3^9} = 3^3 = 27$.So,$3^{y_1}+3^{y_2}+3^{y_3} \geq 81 = 3^4$.Taking $\log_3$ on both sides,$\log_3(3^{y_1}+3^{y_2}+3^{y_3}) \geq 4$,so $m=4$.For $M$,by the $AM$-$GM$ inequality,$\frac{x_1+x_2+x_3}{3} \geq \sqrt[3]{x_1 x_2 x_3}$.Given $x_1+x_2+x_3=9$,we have $3 \geq \sqrt[3]{x_1 x_2 x_3}$,so $x_1 x_2 x_3 \leq 27$.Then $\log_3(x_1 x_2 x_3) = \log_3 x_1 + \log_3 x_2 + \log_3 x_3 \leq \log_3(27) = 3$,so $M=3$.Finally,$\log_2(m^3) + \log_3(M^2) = \log_2(4^3) + \log_3(3^2) = \log_2(64) + 2 = 6 + 2 = 8$.

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