Optimization of Exponential and Logarithmic Expressions

To find the value of $\log_2(m^3) + \log_3(M^2)$, we first need to determine the values of $m$ and $M$.

Step 1: Determine the value of $m$ We are given that $m$ is the minimum possible value of $\log_3(3^{y_1} + 3^{y_2} + 3^{y_3})$ for real numbers $y_1, y_2, y_3$ satisfying $y_1 + y_2 + y_3 = 9$.

By applying the Arithmetic Mean - Geometric Mean (AM-GM) inequality to the positive real numbers $3^{y_1}, 3^{y_2}, 3^{y_3}$, we get: $\frac{3^{y_1} + 3^{y_2} + 3^{y_3}}{3} \geq \sqrt[3]{3^{y_1} \cdot 3^{y_2} \cdot 3^{y_3}}$ $\frac{3^{y_1} + 3^{y_2} + 3^{y_3}}{3} \geq \left(3^{y_1 + y_2 + y_3}\right)^{1/3}$

Substituting $y_1 + y_2 + y_3 = 9$: $\frac{3^{y_1} + 3^{y_2} + 3^{y_3}}{3} \geq \left(3^9\right)^{1/3} = 3^3 = 27$ $3^{y_1} + 3^{y_2} + 3^{y_3} \geq 3 \times 27 = 81$

Equality holds when $y_1 = y_2 = y_3 = 3$. Since $\log_3(x)$ is a strictly increasing function, taking $\log_3$ on both sides gives the minimum value $m$: $m = \log_3(81) = \log_3(3^4) = 4$

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Step 2: Determine the value of $M$ We are given that $M$ is the maximum possible value of $(\log_3 x_1 + \log_3 x_2 + \log_3 x_3) = \log_3(x_1 x_2 x_3)$ for positive real numbers $x_1, x_2, x_3$ satisfying $x_1 + x_2 + x_3 = 9$.

By applying the AM-GM inequality to $x_1, x_2, x_3$: $\frac{x_1 + x_2 + x_3}{3} \geq \sqrt[3]{x_1 x_2 x_3}$

Substituting $x_1 + x_2 + x_3 = 9$: $\frac{9}{3} \geq \sqrt[3]{x_1 x_2 x_3} \implies 3 \geq \sqrt[3]{x_1 x_2 x_3}$ Cubing both sides: $x_1 x_2 x_3 \leq 3^3 = 27$

Equality holds when $x_1 = x_2 = x_3 = 3$. Since $\log_3(x)$ is a strictly increasing function, the maximum value $M$ is: $M = \log_3(27) = \log_3(3^3) = 3$

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Step 3: Evaluate $\log_2(m^3) + \log_3(M^2)$ Using $m = 4$ and $M = 3$:

$\log_2(m^3) = \log_2(4^3) = \log_2((2^2)^3) = \log_2(2^6) = 6$ $\log_3(M^2) = \log_3(3^2) = 2$

Therefore: $\log_2(m^3) + \log_3(M^2) = 6 + 2 = 8$