Question

logx/b-c=log y/c-a=logz/a-b then the value of x^b+c×y^c+a×z^a+b is​

Answer 1

Answer:

let (logx)/(b - c) = (logy)/(c - a) = (logz)/(a -b) = k Then log x = k(b – c), log y = k(c – a), log z = k(a – b) i) log (xyz) = log x + log y + log z) = k(b – c) + k(c – a) + k(a – b) = k(b – c + c – a + a – b) = 0 = log 1 => xyz = 1 ii) log (xa yb zc ) = log xa + log yb + log zc = a log x + b log y + c log z = ak (b – c) + bk (c – a) + ck(a – b) = k[ab – ac + bc – ab + ca – bc] = 0 = log 1 => xa yb zc = 1Read more on Sarthaks.com - https://www.sarthaks.com/489762/if-logx-b-c-logy-c-a-logz-a-b-then-show-that-i-xyz-1-ii-x-a-y-b-z-c-1