Question

If$\left(\frac{\sqrt{7}}{5}\right)^{3x - y} = \frac{875}{2401}$ and \left(\frac{4a}{b}\right)^{6x - y}$$=\left(\frac{2a}{b}\right)^{y - 6x}, for all non-zero real values of a and b,then the value of x+y is

Answer 1

Given the equations:

1. $\left(\sqrt{\frac{7}{3}}\right)^{2x-y} = \frac{875}{2401}$
2. $\left(\frac{4a}{b}\right)^{4x-y} = \left(\frac{2a}{b}\right)^{y-6x}$

For all non-zero real values of 'a' and 'b', we are asked to find the value of x+y.
Let's solve these equations step by step:

1. $\left(\sqrt{\frac{7}{3}}\right)^{2x-y} = \frac{875}{2401}$
   Simplifying the right side, we have $\frac{875}{2401} = \frac{5}{7}$.
   Taking square root of both sides, we get $\sqrt{\frac{7}{3}}^{2x-y} = \frac{5}{7}$.
   Comparing the exponents, we have 2x-y = -2, which gives y = 2x + 2.

2. $\left(\frac{4a}{b}\right)^{4x-y} = \left(\frac{2a}{b}\right)^{y-6x}$
   Dividing both sides by $\left(\frac{2a}{b}\right)^{4x-y}$, we get $\left(\frac{2a}{b}\right)^{2y} = 1$.
   This implies that $\frac{2a}{b} = 1$ (assuming $\frac{2a}{b} \neq 1$ would lead to 6x-y = 0.
   Solving for 'b', we get b = 2a.

Now, we have two equations:
1. y = 2x + 2
2. b = 2a
Substituting the value of 'b' in terms of 'a' from the second equation into the first equation, we get:
y = 2x + 2 (since b = 2a)
From the given information, we can solve for 'x' and 'y':
1. y = 2x + 2
2. y = 6x (assuming $\frac{2a}{b} \neq 1$)

Solving the system of equations, we get x = 2 and y = 12.
Thus, x + y = 2 + 12 = 14.