Question

if log b/log a=3/2,log d/log c=5/4,a-c=9,b-d=?

Answer 1

Assuming a = x^2, and c = y^4, which implies that,

x^2 - y^4 = 9. After factorizing it, we get

(x - y^2)(x + y^2) = 9.

9 = 1 * 9 = 3 * 3 = - 3 * - 3 = - 1 * - 9

So we have 4 cases.

Taking our first case when,

x - y^2 = 1, and x + y^2 = 9. Which implies that,

x = 5. y = +2 or - 2.

a = x^2 = 25 and c = y^4 = 16.

b = a^(3/2) = 125. And d = c^(5/4) = 32.

Therefore b - d = 93.

Second case, when

x - y^2 = 3, and x + y^2 = 3. Which implies that,

x = 3. y = 0.

a = x^2 = 9. c = y^4 = 0.

b = a^(3/2) = 27 And d = c^(5/4) = 0.

Therefore b - d = 27.

Considering our third case when,

x - y^2 = - 3, and x + y^2 = - 3. Which implies that,

x = - 3. y = 0.

a = x^2 = 9. c = y^4 = 0.

b = a^(3/2) = - 27. d = c^(5/4) = 0.

Therefore b - d = - 27.

And coming to our last case,

x - y^2 = - 9 And x + y^2 = - 1. Which implies that,

x = - 5. y = 2 or - 2.

a = x^2 = 25. c = y^4 = 16.

b = a^(3/2) = - 125. d = c^(5/4) = - 32.

Therefore b - d = - 93.
But here's the catch. These are only the integral solutions to this question. We have,

(x^2 - y^4) = 9. So x^2 can be equal to 12, and y^4 = 3. Question being wouldn't there be any problems? And answer is not at all. Try it out for yourself. And we can say, that x^2 = 12 and y^4 = 3 is also a valid solution. a = 12, c = 3.

So value of b - d = a^(3/2) - c^(5/4) = x^3 - y^5

= 12^(3/2) - 3^(5/4). Now you'll have to consider these cases too, for the possible solutions.