Discussion

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

Explanation:

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

⇒ ${\left(\frac{7}{5}\right)}^{\frac{3x-y}{2}}$ = $\frac{125\times 7}{7\times 7\times 7\times 7}$ = ${\left(\frac{5}{7}\right)}^3$ = ${\left(\frac{7}{5}\right)}^{-3}$

⇒ 3x - y = - 6   

⇒ y = 3x + 6   ...(1)

 

Also, ${\left(\frac{4a}{b}\right)}^{6x-y}$ = ${\left(\frac{2a}{b}\right)}^{y-6x}$ = ${\left(\frac{b}{2a}\right)}^{6x-y}$

⇒ ${\left(\frac{4\mathrm{a}}{\mathrm{b}}\times \frac{2\mathrm{a}}{\mathrm{b}}\right)}^{6\mathrm{x}-\mathrm{y}}$ = 1

⇒ ${\left(\frac{8{\mathrm{a}}^2}{{\mathrm{b}}^2}\right)}^{6\mathrm{x}-\mathrm{y}}$ = 1

This is true for all values of a and b. This is only possible when 6x - y = 0
⇒ y = 6x   ...(2)

Solving (1) and (2), we get x = 2 and y = 12.

⇒ x + y = 2 + 12 = 14.

Hence, 14.

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