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

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It’s ($\sqrt{7/5}$)$^{3x-y}$ and not ($\sqrt{7/3}$)$^{2x-y}$. Otherwise we can’t solve this.

And also it’s (4a/b)$^{6x-y}$ and not (4a/b)$^{4x-y}$.

1st equation :- ($\sqrt{7/5}$)$^{3x-y}$ = 875/2401 = 125/343 = 5$^{3}$/7$^{3}$

=> (7/5)$^{(3x-y)/2}$ = (7/5)$^{-3}$  => (3x – y)/2 = -3  => y = 3x + 6

2nd equation :- (4a/b)$^{6x-y}$ = (2a/b)$^{y-6x}$  => 2$^{12x-2y}$ x a$^{6x-y}$ x b$^{-6x+y}$ = 2$^{y-6x}$ x a$^{y-6x}$ x b$^{6x-y}$

=> 2$^{18x-3y}$ x a$^{12x-2y}$ = b$^{12x-2y}$  => (8a$^{2}$/b$^{2}$)$^{6x-y}$ = 1

In this equation if (8a$^{2}$/b$^{2}$) != 1, then we can say that 6x-y=0. Since the only info given for a & b is that it is non zero real number and from this we can’t derive that (8a$^{2}$/b$^{2}$) != 1. But for solving this question we need another equation in terms of x & y. So, assume (8a$^{2}$/b$^{2}$) != 1. Then we can say y = 6x.

After solving y = 3x + 6 and y = 6x, we get x = 2 and y = 12.

Ans is x+y=14.

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