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 $\left( {{{\log }_3}{x_1} + {{\log }_3}{x_2} + {{\log }_3}{x_3}} \right)$ , where x1, x2, x3 are real numbers for which ${x_1} + {x_2} + {x_3} = 9$ . Then find the value of ${\log _2}{\left( m \right)^3} + {\log _3}{\left( M \right)^2}$ .

Answer 1

Hint: Here in this question we use the concept of Arithmetic Mean (AM) and Geometric Mean (GM). We know that GM is always lesser than or equal to AM. We use this concept twice in the solution, first to find the value of m and then to find the value of M. While solving the question we use some logarithmic identities like $\log (x) + \log (y) = \log (xy)$ .

Complete step by step solution: 
We know that, Arithmetic Mean i.e. $AM = \dfrac{{sum\,of\,entries}}{{number\,of\,entries}}$
and Geometric Mean i.e. $GM = {\left( {product\,of\,entries} \right)^{\dfrac{1}{{number\,of\,entries}}}}$
Now, $AM \geqslant GM$ ...(1)
Here $AM = \dfrac{{{3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}}}}{3}$
And $GM = {\left( {{3^{{y_1}}} \times {3^{{y_2}}} \times {3^{{y_3}}}} \right)^{\dfrac{1}{3}}}$
By substituting these values in equation (1), we get
$\dfrac{{{3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}}}}{3} \geqslant {\left( {{3^{{y_1}}} \times {3^{{y_2}}} \times {3^{{y_3}}}} \right)^{\dfrac{1}{3}}}$ ...(2)
We know that, $\left( {{x^a} \times {x^b} \times {x^c}} \right) = {x^{\left( {a + b + c} \right)}}$
This implies that $\left( {{3^{{y_1}}} \times {3^{{y_2}}} \times {3^{{y_3}}}} \right) = {3^{\left( {{y_1} + {y_2} + {y_3}} \right)}}$
Putting this in equation (2), we get
\[\dfrac{{{3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}}}}{3} \geqslant {\left( {{3^{\left( {{y_1} + {y_2} + {y_3}} \right)}}} \right)^{\dfrac{1}{3}}}\]
It is given in the question that ${y_1} + {y_2} + {y_3} = 9$
Therefore, \[{3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}} \geqslant 3 \times {\left( {{3^9}} \right)^{\dfrac{1}{3}}}\]
\[{3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}} \geqslant 3 \times \left( {{3^3}} \right)\]
Thus we get that \[{3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}} \geqslant 81\] ...(3)
It is given that m is the minimum possible value of ${\log _3}({3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}})$ .
This implies that $m = {\log _3}({3^{{y_1}}} + {3^{{y_2}}} + {3^{{y_3}}})$
 $m = {\log _3}(81)$
$81$ can be written as ${3^4}$ .
So, $m = {\log _3}\left( {{3^4}} \right)$
 $m = 4{\log _3}\left( 3 \right)$
Also, ${\log _3}(3) = 1$
Thus $m = 4$ ...(4)
Similarly we find the value of M.
$AM \geqslant GM$
This time $AM = \dfrac{{{3^{{x_1}}} + {3^{{x_2}}} + {3^{{x_3}}}}}{3}$
And \[GM = {\left( {{3^{{x_1}}} \times {3^{{x_2}}} \times {3^{{x_3}}}} \right)^{\dfrac{1}{3}}}\]
Again substituting these equations in (1), we get
\[\dfrac{{{x_1} + {x_2} + {x_3}}}{3} = {\left( {{x_1} \times {x_2} \times {x_3}} \right)^{\dfrac{1}{3}}}\]
Given: ${x_1} + {x_2} + {x_3} = 9$
Therefore, $\dfrac{9}{3} = {\left( {{x_1} \times {x_2} \times {x_3}} \right)^{\dfrac{1}{3}}}$
 $3 = {\left( {{x_1} \times {x_2} \times {x_3}} \right)^{\dfrac{1}{3}}}$
This implies that ${3^3} = \left( {{x_1} \times {x_2} \times {x_3}} \right)$
 $27 = \left( {{x_1} \times {x_2} \times {x_3}} \right)$
$\left( {{x_1} \times {x_2} \times {x_3}} \right) = 27$ ...(5)
Now it is given in the question that M is the maximum possible value of $\left( {{{\log }_3}{x_1} + {{\log }_3}{x_2} + {{\log }_3}{x_3}} \right)$ .
So, $M = \left( {{{\log }_3}{x_1} + {{\log }_3}{x_2} + {{\log }_3}{x_3}} \right)$
Also, we know that $\log (x) + \log (y) = \log (xy)$ .
So we write $M = \log ({x_1} \times {x_2} \times {x_3})$
Substituting from the required value from equation (5), we get
 $M = {\log _3}27$
Thus, $M = 3$ ...(6)
Now we find the value of ${\log _2}{\left( m \right)^3} + {\log _3}{\left( M \right)^2}$ .
Substitute the value of m and M from the equations (4) and (5).
 ${\log _2}{\left( m \right)^3} + {\log _3}{\left( M \right)^2} = {\log _2}{\left( 4 \right)^3} + {\log _3}{\left( 3 \right)^2}$
 ${\log _2}{\left( m \right)^3} + {\log _3}{\left( M \right)^2} = {\log _2}{\left( {{2^2}} \right)^3} + {\log _3}{\left( 3 \right)^2}$
On solving further we get,
${\log _2}{\left( m \right)^3} + {\log _3}{\left( M \right)^2} = {\log _2}({2^6}) + {\log _3}{\left( 3 \right)^2}$
This implies ${\log _2}{\left( m \right)^3} + {\log _3}{\left( M \right)^2} = 6 \times {\log _2}2 + 2 \times {\log _3}3$
And we know that ${\log _a}(a) = 1$
So we get, ${\log _2}{\left( m \right)^3} + {\log _3}{\left( M \right)^2} = (6 \times 1) + (2 \times 1)$
Thus we get the final answer, ${\log _2}{\left( m \right)^3} + {\log _3}{\left( M \right)^2} = 8$ .

Note: While solving the question do not write $\log (x + y) = \log x + \log y$ as this is not a logarithmic identity instead $\log (x) + \log (y) = \log (xy)$ is a logarithmic identity. Also we should always keep in mind the base of log given in the question because it may not be to the base e (natural log) or to the base 10 always.