Question

Prove that: (2^a/2^b)^a+b × (2^b/2^c)^b+c × (2^c/2^a)^c+a = 1

Answer 1

Solution :

LHS =

(2^a/2^b)^a+b*(2^b/2^c)^b+c*(2^c/2^a)^c+a

=2^(a-b)(a+b)*2^(b-c)(b+c)*2^(c-a)(c+a)

**************************************
We know the Exponential laws:

i ) (x^m/x^n) = x^m-n

ii ) x^m × x^n = x^m+n

****************************************
= 2^a²-b² * 2^b²-c² * 2^c²-a²

= 2^a²-b²+b²-c²+c²-a²

= 2^0

= 1

= RHS

•••••

Answer 2

plzz refer the attachment