Solve this problem : If (Log x)/(b-c)=(Log y)/(c-a)=(Log z)/(a-b) Prove that x^(b+c)*y^(c+a)*z^(b+a)=1

We'll take logarithms both
sides:

log [x^(b+c)*y^(c+a)*z^(b+a)] = log 1
(*)

We'll transform the logarithm of the product into a
sum:

log [x^(b+c)*y^(c+a)*z^(b+a)] = log x^(b+c) +
logy^(c+a) + log z^(b+a)

We'll use the power property of
logarithms:

log x^(b+c) = (b+c)*log
x

logy^(c+a) = (c+a)*log y

log
z^(b+a) = (b+a)*log z

The relation (*) will become
(1):

(b+c)*log x + (c+a)*log y + (b+a)*log z =
0  (1)

But, from enunciation, we'll
have:

(Log x)/(b-c) = (Log y)/(c-a) = (Log
z)/(a-b)

We'll cross
multiply:

(Log x)*(c-a) = (Log
y)*(b-c)

Log y = (Log
x)*(c-a)/(b-c)

Log z = (Log
x)*(a-b)/(b-c)

So, the relation (1) will become
(2):

(b+c)*Log x + (c+a)*(Log x)*(c-a)/(b-c) + (b+a)*(Log
x)*(a-b)/(b-c) = 0 

We'll factorize by log
x:

log x [b+c + (c^2-a^2)/(b-c) + (a^2 - b^2)/(b-c)] =
0

We'll divide by log x and we'll remove the
brackets:

b^2 - c^2 + c^2-a^2 + a^2 - b^2 =
0

We'll remove like terms and we'll
get:

0 =
0

So,
x^(b+c)*y^(c+a)*z^(b+a)=1 if and only if (Log x)/(b-c) = (Log y)/(c-a) = (Log
z)/(a-b).