Question

If a = b²ˣ, b = c²ʸ and c = a²ᶻ, prove that xyz = 1/8.

Answer 1

Given ,

a = b²ˣ, b = c²ˣ and c = a²ᶻ

Solving a = b²ˣ

Put value of b in a =b²ˣ, we get

=> a = (c²ʸ)²ˣ [given b = b²ˣ ]

=> a = (c)²ˣ²ʸ

Now , put the value of c in [a = (c)²ˣ²ʸ], we get

=> a = ( a²ᶻ )²ˣ²ʸ [ given c = a²ᶻ ]

=> a = (a)²ˣ²ʸ²ᶻ

Now , we have the same base on both sides of the equation , so the power on both sides can be compared

=> 8xyz = 1

=> xyz = 1/8

Hence , proved