Plz prove it fast. If 2^a=3^b=6^c then show that c = ab/a+b
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Let 2^a = 3^b = 6^c = k.
2 = k^1/a ------------------- (1)
3 = k^1/b --------------------- (2)
6 = k^1/c ---------------------(3).
We know that 6 = 2 * 3.
So,
k^1/c = k^1/a * k^1/b
k^1/c = k^(1/a + 1/b)
1/c = 1/a + 1/b
1/c = (a + b)/ab
c = ab/(a+b).
LHS = RHS.
Hope this helps!
2 = k^1/a ------------------- (1)
3 = k^1/b --------------------- (2)
6 = k^1/c ---------------------(3).
We know that 6 = 2 * 3.
So,
k^1/c = k^1/a * k^1/b
k^1/c = k^(1/a + 1/b)
1/c = 1/a + 1/b
1/c = (a + b)/ab
c = ab/(a+b).
LHS = RHS.
Hope this helps!
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