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=d
so,2^a=d
=)d^1/a=2----1) and 3^b=d
=)d^1/b=3------2)and 6^c=d
=)d^1/c=6------3)
Since, 6=2×3-----4)
Now put value of 2), 3) and 6) from from eq 1), 2) , 3)
we get :-
d^1/a × d^1/b = d^1/c
d(1/a=1/b) = d^1/c
=) 1/c = 1/a + 1/b ⇒1/c = (b+a)/(ab)
=)c=ab / a+b
Hope it helps
Let 2^a=3^b=6^c=d
so,2^a=d
=)d^1/a=2----1) and 3^b=d
=)d^1/b=3------2)and 6^c=d
=)d^1/c=6------3)
Since, 6=2×3-----4)
Now put value of 2), 3) and 6) from from eq 1), 2) , 3)
we get :-
d^1/a × d^1/b = d^1/c
d(1/a=1/b) = d^1/c
=) 1/c = 1/a + 1/b ⇒1/c = (b+a)/(ab)
=)c=ab / a+b
Hope it helps
nitthesh7:
It is wrong sis pls change it
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