Question

if x^y^z=2^8 then find the maximum possible value of x×y×z where x,y,z >0
please solve ....…..​

Answer 1

Given:

x^y^z=2^8 and  x,y,z >0

To find:

If x^y^z=2^8 then find the maximum possible value of x×y×z where x,y,z >0

Solution:

From given, we have,

x^y^z = 2^8

⇒ x = 2

∴ x = 2 ....(1)

⇒ y^z = 8

there is only one possibility, where y^z = 8

therefore, we have,

y^z = 8 = 2^3

∴ y = 2 .....(2)

∴ z = 3 ......(3)

Using the values (1), (2) and (3), we get the value of x×y×z  as,

x × y × z  = 2 × 2 × 3 = 12

12 > 0

Therefore,  the maximum possible value of x × y × z where x, y, z >0 is 12.