Math, asked by bhartibaghel264, 11 months ago

Form the partial differential e
11. z=(x + a) (y + b)​

Answers

Answered by piyushsahu624
0

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.

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