Question

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

Answer 1

Let 2^a=3^b=6^c=x

Then

2^a=x , 3^b=x , 6^c=x

2=x^1/a , 3=x^1/b , 6=x^1/c

We know that

2*3=6

Then

x^1/a * x^1/b = x^1/c

x^1/a+1/b=x^1/c

1/a+1/b=1/c

a+b/ab=1/c

c=ab/a+b

hence prooved

Answer 2

Given,

$2^{a}$ = $3^{b}$ = $6^{c}$

To Find,

Show that c =ab / (a+b)

Solution,

Let $2^{a}$ = $3^{b}$ = $6^{c}$ = k

2 = $k^{(1/a)}$, 3 = $k^{(1/b)}$ and 6 = $k^{(1/c)}$

2*3=6

$k^{(1/a)}$ * $k^{(1/b)}$ =  $k^{(1/c)}$

$k^{(1/a +a/b)}$ = $k^{(1/c)}$

1/a +a/b = 1/c

(a+b)/ab = 1/c

c= ab/(a+b)

Hence, c =ab/a+b.