Question

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

Answer 1

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

$i ) 2^{a} = k \implies 2 = k^{\frac{1}{a}}\:--(1)$

$ii ) 3^{b} = k \implies 3 = k^{\frac{1}{b}}\:--(2)$

$iii ) k = 6^{c}$

$\implies k^{\frac{1}{c}} = 6$

$\implies k^{\frac{1}{c}} = 2 \times 3\\= k^{\frac{1}{a}} \times k^{\frac{1}{b}}\: [ From \: (1) \:and \: (2) ]$

$=k^{\frac{1}{a} + \frac{1}{b}}$

$\implies \frac{1}{c} = \frac{1}{a} + \frac{1}{b}$

$\boxed { \pink { If \: a^{m} = a^{n} \implies m = n }}$

$\implies \frac{1}{c} = \frac{b +a}{ab}$

$\implies \frac{ab}{a+b} = c$

$\red {2^{a} = 3^{b} = 6^{c}} \\\green { \implies c = \frac{ab}{a+b}}$

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Answer 2

Hey mate !! Here you go :-

Given that, $2^{a} =3^{b} =6^{c}$

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

Before you solve see this also this will help you to understand better :-

$2^{2} =4$

$2=4^{1/2}$

$2=2$

Now,

$2^{a} =k$

$2=k^{1/a}$-- (1)

$3^{b} =k$

$3=k^{1/b}$-- (2)

$6^{c} =k$

$6=k^{1/c}$-- (3)

As we know that,

2*3=6

Putting values of 2,3 and 6

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

Bases are same so powers are equal

$\frac{1}{a} *\frac{1}{b} =\frac{1}{c}$

Take LCM of a and b that is ab

$\frac{a+b}{ab} =\frac{1}{c}$

Cross multiply to get c

$c=\frac{ab}{a+b}$

Hence proved...

Hope it is helpful..

Bye !!