If $$(\sqrt{\frac{7}{5}})^{3x-y}=\frac{875}{2401}$$ and $$(\frac{4a}{b})^{6x-y}=(\frac{2a}{b})^{y-6x}$$, for all non-zero real values of a and b, then the value of $$x+y$$ is
Correct Answer: 14
$$(\sqrt{\frac{7}{5}})^{3x-y}=\frac{875}{2401}$$
$$\left(\frac{7}{5}\right)^{\frac{\left(3x-y\right)}{2}}=\frac{125}{343}$$
$$\left(\frac{7}{5}\right)^{\frac{\left(3x-y\right)}{2}}=\left(\frac{7}{5}\right)^{-3}$$
3x-y = -6
$$(\frac{4a}{b})^{6x-y}=(\frac{2a}{b})^{y-6x}$$
Therefor, y=6x as the bases are different so the power should be zero for the results to be equal.
3x-y=-6
or, 3x - 6x = -6
or x= 2
y= 6x = 12
x+y = 14
Create a FREE account and get: