2^a=3^b=6^c then show that c=ab/a+b
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We have 2^a=3^b=6^c
= 2^c *3^c --(1)
This gives 2^c*3^c=3^b
⇒ 2^c= 3^(b-c)
and 2^c*3^c= 2^a
⇒ 3^c= 2^(a-c)
substituting value in one we have 2^c*3^c= 3^(b-c)*3^(a-c)
then we have c= b-c
⇒ b=2c
and c=a-c
⇒ a = 2c
RHS= ab/(a+b)= 2c*2c/( 2c+2c)=c=LHS
= 2^c *3^c --(1)
This gives 2^c*3^c=3^b
⇒ 2^c= 3^(b-c)
and 2^c*3^c= 2^a
⇒ 3^c= 2^(a-c)
substituting value in one we have 2^c*3^c= 3^(b-c)*3^(a-c)
then we have c= b-c
⇒ b=2c
and c=a-c
⇒ a = 2c
RHS= ab/(a+b)= 2c*2c/( 2c+2c)=c=LHS
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