Question

log2/b-c=log3/c-a=log5/a-b then 2power a*3power b*5 power c=?​

Answer 1

Given :  log 2/ (b - c)  = log 3/ (c - a)  = log 5 / (a - b)  

To find :  2power a*3power b*5 power c

Solution:

Let say

log 2/ (b - c)  = log 3/ (c - a)  = log 5 / (a - b)  = k

=> log 2 = k(b - c)  =>  $10^{ k(b - c)} =2$  

   log 3 = k (c - a)  =>  $10^{ k(c - a)} =3$

   log 5 = k (a - b)  =>  $10^{ k(a - b)} =5$

$10^{ k(b - c)} =2$    =>  $10^{ ak(b - c)} =2^a$  

$10^{ k(c - a)} =3$ =>  $10^{ bk(c - a)} =3^b$

$10^{ k(a - b)} =5$ => $10^{ ck(a - b)} =5^c$

$2^a.3^b.5^c=$    $10^{ ak(b - c)}$   $10^{ bk(c - a)}$ $10^{ ck(a - b)}$

=> $2^a.3^b.5^c=$  10⁰

=> $2^a.3^b.5^c=$  1

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