Logarithms
Logarithmic Functions
MJAT None
Grade 12
Question:
Let $m$ be the minimum possible value of $\log_3(3^{y_1} + 3^{y_2} + 3^{y_3})$, where $y_1, y_2, y_3$ are real numbers for which $y_1 + y_2 + y_3 = 9$. Let $M$ be the maximum possible value of $(\log_3 x_1 + \log_3 x_2 + \log_3 x_3)$, where $x_1, x_2, x_3$ are positive real numbers for which $x_1 + x_2 + x_3 = 9$. Then the value of $\log_2(m^3) + \log_3(M^2)$ is _____
Step-by-Step Solution
Key Concept: Use the fact that the first sum is fixed (so the log is constant) and apply AM‑GM to maximize the product under a fixed sum, converting the sum of logs to a log of a product.
1. **Find the minimum value $m$**:
Using the AM-GM inequality on $3^{y_1}, 3^{y_2}, 3^{y_3}$:
$$\frac{3^{y_1} + 3^{y_2} + 3^{y_3}}{3} \ge \sqrt[3]{3^{y_1 + y_2 + y_3}} = \sqrt[3]{3^9} = 3^3 = 27$$
So:
$$3^{y_1} + 3^{y_2} + 3^{y_3} \ge 3 \times 27 = 81$$
Therefore:
$$\log_3(3^{y_1} + 3^{y_2} + 3^{y_3}) \ge \log_3(81) = 4$$
This minimum occurs when $y_1 = y_2 = y_3 = 3$, giving $m = 4$.
2. **Find the maximum value $M$**:
Using the AM-GM inequality on positive real numbers $x_1, x_2, x_3$:
$$\frac{x_1 + x_2 + x_3}{3} \ge \sqrt[3]{x_1 x_2 x_3} \implies \frac{9}{3} \ge \sqrt[3]{x_1 x_2 x_3} \implies x_1 x_2 x_3 \le 27$$
The expression $(\log_3 x_1 + \log_3 x_2 + \log_3 x_3) = \log_3(x_1 x_2 x_3)$. Thus:
$$\log_3(x_1 x_2 x_3) \le \log_3(27) = 3$$
This maximum occurs when $x_1 = x_2 = x_3 = 3$, giving $M = 3$.
3. **Compute the final expression**:
$$\log_2(m^3) + \log_3(M^2) = \log_2(4^3) + \log_3(3^2) = \log_2(64) + \log_3(9) = 6 + 2 = 8.$$
Correct Answer: 8