Question

If log x/b-c = log y/c-a = log z/a-b, then which of the following is/are correct :-
a) xyz=1
b) (x^a)(y^c)(z^c)=1
c) (x^b+c)(y^c+a)(z^a+b)=1
d) xyz=(x^a)(y^b)(z^c)

Ans. All are correct.
But how?

Answer 1

here,

logx/( b - c) =logy/( c - a) = logz/(a - b) =K

logx = K(b - c) -----(1)
logy = K(c - a) ------(2)
logz = K(a - b) --------(3)

add eqns (1) (2) and (3)

logx + logy + logz = K( b - c +c -a + a - b)

logxyz = K.0 = 0

xyz = 1 hence, option (a) is correct .

now,
logx = K(b - c)
alogx = Ka(b - c)
logx^a = K( ab - ac)

similarly ,

blogy = 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 hence, option (b) is also correct .

logx = K(b - c)
(b+ c)logx = K(b - c)(b + c) = K(b² - c²)

similarly,

logy = K(c - a)
(c + a)logy = K(c² - a²)

logz = K(a - b)
(a + b)logz = K( a² - b²)

add all terms ,

log{(x^(b+c)).( y^(c + a)) .(z^(a + b)) } = K(b² - c² + c² - a² + a² - b² } = k×0 = 0

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

hence, option (C) is also correct .

xyz = 1
x^a.y^b.z^c = 1

so, xyz = x^a.y^b.z^c
hence, option (d) is correct .


so, here all options (A), (B) , (C) , (D) are correct .