Question

if 2^a = 3^b = 6^c then show that c = ab/a+b

plz solve in a simple way.. according to 9th grade..

Answer 1

Hello Mate!

Let's assume that all terms are equal to k.

2^a = 3^b = 6^c = k

2 = k^(1/a)

3= k^(1/b)

6 = k^(1/c)

2 × 3 = 6

k^(1/a) × k^(1/b) = k^(1/c)

Since exponents add when same bases multiplying.

1 / a + 1 / b = 1 / c

( b + a ) / ab = 1 / c

c( a + b ) = ab

c = ab / ( a + b )

Have great future ahead!

Answer 2

Answer:

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)=