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 $({\log }_{3}1+{\log }_{3}2+{\log }_{3}3),$ where$x_{1,2,3}$ are positive real numbers for which $x_1+x_2+x_3=9.$Then the value of $lo{g}_2(3)+{\log }_3(2)$ is _____

Answer 1

The value of $lo{g}_2(3)+{\log }_3(2)$ is $8$

Explanation for the correct answer:

Finding the value of $lo{g}_2(3)+{\log }_3(2)$

Given: $m$ is 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 $y_1+y_2+y_3=9$.

and M is the maximum possible value of $({\log }_{3}1+{\log }_{3}2+{\log }_{3}3),$ where$x_1,x_2,x_3$ are positive real numbers for which $x_1+x_2+x_3=9.$

$$\begin{gathered} \frac{(3^{y_1}+3^{y_2}+3^{y_3})}{3}\ge {(3^{y_1}+3^{y_2}+3^{y_3})}^{\frac{1}{3}} \\ (3^{y_1}+3^{y_2}+3^{y_3})\ge 3.{\left(3^9\right)}^{{}^{\frac{1}{3}}} \\ (3^{y_1}+3^{y_2}+3^{y_3})\ge 81 \\ l{\mathrm{og}}_3(3^{y_1}+3^{y_2}+3^{y_3})\ge 4 \end{gathered}$$

So, it shows minimum value of $m=4$

Now,

$$\begin{gathered} \frac{x_1+x_2+x_3}{3}\ge {(x_1.x_2.x_3)}^{\frac{1}{3}} \\ {\left(\frac{9}{3}\right)}^3>x_1{x}_2{x}_3 \\ \log x_1{x}_2{x}_3\le {\log }_{3}27 \\ {\log }_3{x}_1+{\log }_3{x}_2+{\log }_3{x}_3\le 3 \end{gathered}$$

So, it shows the maximum value of $M=3$

Now,

$$\begin{gathered} {\log }_2{m}^3+{\log }_3{M}^2 \\ {\log }_2{4}^3+{\log }_3{3}^2 \\ =6+2 \\ =8 \end{gathered}$$

Hence the value of $lo{g}_2(3)+{\log }_3(2)$ is $8$