Question

2^a=3^b=6^c prove that c=ab/a+b

Answer 1

Let 2^a=3^b=6^c=K (K≠0)

2^a = K ⇒2 = K^1/a

3^b = K ⇒3 = K^1/b

6^c = K ⇒6 = K^1/c

We know that:

6 = 2 * 3

⇒ K^1/c = K^1/a * K^1/b

⇒ K^1/c = K^1/a+1/b

⇒ K^1/c = K^a+b/ab

⇒ 1/c = a+b/ab [∵ the bases are equal, equating them:]

⇒ c = ab/a+b