Question

if 2a=3b=6c then show that c= ab/a+b
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Answer 1

Answer:

2a=3b=6c then c=ab/a+b

c=2×3/2+3

c=6/5 c=1.2

Answer 2

QUESTION:

• If 2^a = 3^b = 6^c, show that c = ab/(a + b)

ANSWER:

Given:

• $2^a=3^b=6^c$

To Prove:

• c = ab/(a + b)

Proof:

We are given that,

$\implies\sf 2^a=3^b=6^c$

Let these be equal to a constant k. That is,

$\implies\sf 2^a=3^b=6^c=k$

So,

$\implies\sf 2^a=k$

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

Similarly,

$\implies\sf 3^b=k$

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

And,

$\implies\sf 6^c=k$

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

We can write it as,

$\implies\sf (2\times3)=k^{\frac{1}{c}}$

Substituting values of 2 and 3 from (1) and (2),

$\implies\sf (k^{\frac{1}{a}}\times k^{\frac{1}{b}})=k^{\frac{1}{c}}$

So,

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

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

As, the bases are same(k), we compare the powers,

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

So,

$\implies\bf c=\dfrac{a+b}{a+b}$

HENCE PROVED!!!