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