Question

prove that fast.....

Answer 1

solved with proof.... ..detailed proof.

Answer 2

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

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

Now,

(a)

2^a = k

Apply log on both sides, we get

log(2)^a = log k

alog 2 = log k

a = log k/1og2

$\frac{1}{a} = \frac{1}{ \frac{log k}{log 2} }$

$\frac{1}{a} = \frac{log 2}{log k}$


(b)

3^b = k

Apply log on both sides, we get

log(3)^b = log k

b log 3 = log k

b = log k/log 3

$\frac{1}{b} = \frac{log 3}{log k}$



(c)

6^-c = k

Apply log on both sides, we get

-c log 6 = log k

c = -log k/log 6

$\frac{1}{c} = \frac{-log 6}{log k}$


Therefore:

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} :$

$= \ \textgreater \ \frac{log 2}{log k} + \frac{log 3}{log k} - \frac{log 6}{log k}$

0.


Hope this helps!