Question

if log x/b-c = log y /c-a = log z /a-b than prove that

(I) x^b+c .y^c+a .z^a+b=1
(2)x^a.y^b.z^c=1

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Answer 1

Answer:

1) :-

now,

logx= K(b-c)

alogx=Ka (b-c)

logx^a= K (ab-ac)

similarly,

blog y= Kb(c-a) =K (bc-ab)

logy^a=K(bc-ab)

clogz= K(ac-bc)

logz^c = K(ac-bc)

add all terms,

logx^a+ logy^b + logz^c = K(ab-ac+bc- ab+ ac-bc) = 0

log{(x^a) (y^b) (z^c) }= 0

{(x^a) (y^b) (z^c) }=1 .

Answer 2

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

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