Form the partial differential e
11. z=(x + a) (y + b)
Answers
Answer:
First solution:
We have:
x = y^a => log(x) = alog(y) => a = log(x)/log(y) … (1)
y = z^b => log(y) = blog(z) => b = log(y)/log(z) … (2)
z = x^c => log(z) = clog(x) => c = log(z)/log(x) … (3)
By multiplying (1), (2) and (3) by parts, we obtain:
abc = [(log(x))/log(y)][(log(y))/log(z)][(log(z))/log(x)] => abc = 1
Second solution:
x = y^a = (z^b)^a = z^(ab) = (x^c)^(ab) = x^(abc)
Therefore, since x^1 = x^(abc) => abc = 1
For the case x = 1, we could have any value for abc, but the equation x = 1 will imply log1(z) = c, which is impossible, unless z = 1, since we cannot have logarithm of a number different than 1 with base 1 and finite value.
In the case at which one of x, y, z is equal to 1, we have:
Let x = 1, then by (3) we take z = 1 and by (2) we obtain y = 1. Therefore, then we have x = y = z = 1. For this case, abc could take any real value . The problem is that log1(1) is undefined, which means it can take any real value, but then the expression log(x) will not be a function.
Third solution:
x = y^a … (1)
z^b = y … (2)
By dividing by parts (1) and (2), we take:
x/(z^b) = y^(a-1) … (3)
Now from z = x^c, the equation (3) becomes:
x/[(x^c)^b] = y^(a-1) => x/(x^(bc)) = y^(a-1) => x^(1-bc) = y^(a-1) … (4)
By combining the equations x = y^a and (4), we obtain:
(y^a)^(1-bc) = y^(a-1) => y^(a-abc) = y^(a-1) => a - abc = a - 1 => abc = 1
Fourth solution:
x = y^a … (1)
y = z^b … (2)
z = x^c … (3)
From (1), (2) and (3), we take:
xyz = (y^a)(z^b)(x^c) => log(xyz) = log[(y^a)(z^b)(x^c)] =>
log(x) + log(y) + log(z) = log(y^a) + log(z^b) + log(x^c) =>
log(x) + log(y) + log(z) = alog(y) + blog(z) + clog(x) =>
(c-1)log(x) + (a-1)log(y) + (b-1)log(z) = 0 … (4)
Now, from (1), (2) and (3), we take:
log(x) = alog(y) … (5)
log(y) = blog(z) … (6)
log(z) = clog(x) … (7)
Therefore, by combining equations (4), (5), (6) and (7), we have:
(c-1)log(x) + (a-1)log(y) + (b-1)log(z) = 0 =>
(c-1)alog(y) + (a-1)blog(z) + (b-1)clog(x) = 0 … (8)
By substituting y = z^b and x = z^(1/c), which was taken from (2) and (3) to (8), we take:
(c-1)alog(z^b) + (a-1)blog(z) + (b-1)clog(z^(1/c)) = 0 =>
[(c-1)a]blog(z) + [(a-1)b]log(z) + [(b-1)(1/c)c]log(z) = 0 =>
[ab(c-1) + (a-1)b + (b-1)]logz = 0 … (9)
Now, we discriminate two cases:
First case: log(z) = 0 => z = 1
Now if z = 1, from (2) it follows that:
log1(y) = b, which is impossible, since we cannot have logarithm with base 1 and finite value.
Second case: log(z) is different than 0. Then, from equation (9), we take:
[ab(c-1) + (a-1)b + (b-1)] = 0 => abc - ab + ab - b +b - 1 = 0 => abc = 1 !!!
In the case at which one of x, y, z is equal to 1, we have:
Let x = 1, then by (3) we take z = 1 and by (2) we obtain y = 1. Therefore, then we have x = y = z = 1. For this case, abc could take any real value . The problem is that log1(1) is undefined, which means it can take any real value, but then the expression log(x) will not be a function.