Question

if 2
${2}^{a \: } = {3}^{b} = {6}^{ - c} \\ prove \: that \: \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$
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Answer 1

★ Given -:

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

★ To prove -:

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$

★ Solution -:

Assume

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

$2^a = K ; 2 = K ^{\frac{1}{a}}$

(by multiplying 1/a to the powers on both sides)

$3^b = K , 3 = K ^{\frac{1}{b}}$

$6^{-c} = K , 6 = K ^{- \frac{1}{c}}$

We know that ,

6 = 2 × 3

by putting the values in terms of K

$K^{-\frac{1}{c}} = K^{\frac{1}{a}} \times K^{\frac{1}{b}}$

by applying the laws of exponents

$a^m \times a^n = a^{m+n}$

$K^{-\frac{1}{c}} = K^{\frac{1}{a} + \frac{1}{b}}$

we know that, when the base is same power also becomes equal.

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

by transposing

$\purple{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0}$

→ Hence proved .

Answer 2

$2^a = 3^b = 6^{-c} = K2 </p><p>a</p><p> =3 </p><p>b</p><p> =6 </p><p>−c</p><p> =K</p><p>$

$2^a = K ; 2 = K ^{\frac{1}{a}}2 </p><p>a</p><p> =K;2=K </p><p>a</p><p>1</p><p> </p><p>$

(by multiplying 1/a to the powers on both sides)

$3^b = K , 3 = K ^{\frac{1}{b}}3 </p><p>b</p><p> =K,3=K </p><p>b</p><p>1</p><p>$

$6^{-c} = K , 6 = K ^{- \frac{1}{c}}6 </p><p>−c</p><p> =K,6=K </p><p>− </p><p>c</p><p>1</p><p> </p><p> </p><p>$

We know that ,

6 = 2 × 3

Let the values be in terms of K

$K^{-\frac{1}{c}} = K^{\frac{1}{a}} \times K^{\frac{1}{b}}K </p><p>− </p><p>c</p><p>1</p><p> </p><p> </p><p> =K </p><p>a</p><p>1</p><p> </p><p> </p><p> ×K </p><p>b</p><p>1</p><p> </p><p> </p><p>$

by applying the laws of exponents

$</p><p>a^m \times a^n = a^{m+n}a </p><p>m</p><p> ×a </p><p>n</p><p> =a </p><p>m+n</p><p>$

$</p><p>K^{-\frac{1}{c}} = K^{\frac{1}{a} + \frac{1}{b}}K </p><p>− </p><p>c</p><p>1</p><p> </p><p> </p><p> =K </p><p>a</p><p>1</p><p> </p><p> + </p><p>b</p><p>1</p><p> </p><p> </p><p>$

As we know, when the base is same power also becomes equal.

$-\frac{1}{c} = \frac{1}{a} + \frac{1}{b}− </p><p>c</p><p>1</p><p> </p><p> = </p><p>a</p><p>1</p><p> </p><p> + </p><p>b</p><p>1</p><p>$

by transposing

$\purple{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0} </p><p>a</p><p>1</p><p> </p><p> + </p><p>b</p><p>1</p><p> </p><p> + </p><p>c</p><p>1</p><p> </p><p> =0 (proved)</p><p></p><p>$