Exponential Equations and Variables

Question

If $\left(\sqrt{\frac{7}{5}}\right)^{3x-y} = \frac{875}{2401}$ 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

Accepted Answer

14. From the second equation: $$\left(\frac{4a}{b}ight)^{6x-y} = \left(\frac{2a}{b}ight)^{y-6x} = \left(\frac{b}{2a}ight)^{6x-y}$$ Since this holds for all non-zero $a, b$, the exponent must be zero: $$6x - y = 0 \implies y = 6x$$ Now substitute $y = 6x$ into the first equation: $$\left(\sqrt{\frac{7}{5}}ight)^{3x - 6x} = \left(\frac{7}{5}ight)^{-3x/2}$$ On the RHS: $$\frac{875}{2401} = \frac{125 	imes 7}{343 	imes 7} = \frac{125}{343} = \left(\frac{5}{7}ight)^3 = \left(\frac{7}{5}ight)^{-3}$$ So: $$-\frac{3x}{2} = -3 \implies x = 2$$ Then $y = 6(2) = 12$. Thus $x + y = 2 + 12 = 14$.

Answer

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Exponential Equations and Variables

Explanation

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