Question

if logx/b-c=logy/c-a =log^2 then prove thatx^a y^bz^c=1​

Answer 1

Answer: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

hencrect .