Question

Let X1, X2, X3, X4, X5 are positive numbers such that X1 + x2 + x3 + x4 + X5 = 15. If M is maximum value of (1 + x1)(2 + x2)(3 + x3)(4 + x4)(5
+ X5) then the value of log6 M is
​

Answer 1

5464321 isanswer

Step-by-step explanation:

Answer 2

given sum of five positive numbers $15$, answer the question based on this information

Explanation:

1. given five positive numbers $x_1, x_2,x_3,x_4\ and\ x_5$ such that                               $x_1+x_2+x_3+x_4+x_5=15$   --(i)  
2. substituting $x_1->x_1+1,\ x_2->x_2+2,\ x_3->x_3+3,\ x_4->x_4+4,\ x_5->x_5+5$  in (i) we get , $(x_1+1)+(x_2+2)+(x_3+3)+(x_4+4)+(x_5+5)=30$    ---(ii)        
3. here using Arithmetic - Geometric Mean inequality that is for given positive numbers,  $A.M\geq G.M$ we have,                                                            $->\frac{(x_1+1)+(x_2+2)+(x_3+3)+(x_4+4)+(x_5+5)}{5}\geq \sqrt[5]{(x_1+1)(x_2+2)(x_3+3)(x_4+4)(x_5+5)} \\->\sqrt[5]{(x_1+1)(x_2+2)(x_3+3)(x_4+4)(x_5+5)} \leq \frac{30}{5} \\->{(x_1+1)(x_2+2)(x_3+3)(x_4+4)(x_5+5)} \leq 6^{5}$  
4. given that maximum value of $(x_1+1)(x_2+2)(x_3+3)(x_4+4)(x_5+5)}$ be denoted by M. hence we have,                                                                                      $M=6^5\\\log_{6}M= \log_{6}6^5\\\log_{6}M=5$   -----> ANSWER