Question

logx/y-z=logy/z-x=loz/x-y prove that xpoelwer y+z.ypowerz+x.zpowerx+y=1​

Answer 1

Question:

If logx/(y-z) = logy/(z-x) = logz/(x-y) , then prove that ;

[x^(y+z)]•[y^(z+x)]•[z^(x+y)] = 1.

Note:

• log(a•b) = log(a) + log(b)

• log(a/b) = log(a) - log(b)

• log(a^b) = b•log(a)

Solution:

Let ,

logx/(y-z) = logy/(z-x) = logz/(x-y) = k

where, "k" is any constant.

Thus;

=> logx/(y-z) = k

=> logx = k•(y-z)

=> (y+z)•logx = k•(y-z)•(y+z)

=> log{x^(y+z)} = k•(y^2 - z^2) ----------(1)

Also;

=> logy/(z-x) = k

=> logy = k•(z-x)

=> (z+x)•logy = k•(z-x)•(z+x)

=> log{y^(z+x)} = k•(z^2 - x^2) ----------(2)

Also;

=> logz/(x-y) = k

=> logz = k•(x-y)

=> (x+y)•logz = k•(x-y)•(x+y)

=> log{z^(x+y)} = k•(x^2 - y^2) ----------(3)

Now,

Adding eq-(1) , (2) and (3), we get;

=> log{x^(y+z)} + log{y^(z+x)}

+ log{z^(x+y)} = k•(y^2 - z^2)

+ k•(z^2 - x^2) + k•(x^2 - y^2)

=> log[x^(y+z)]•[y^(z+x)]•[z^(x+y)] = 0

=> [x^(y+z)]•[y^(z+x)]•[z^(x+y)] = e^0

=> [x^(y+z)]•[y^(z+x)]•[z^(x+y)] = 1

Hence proved .