Question

If $$16\left(\frac{a - x}{a + x}\right)^3 = \left(\frac{a + x}{a - x}\right)$$, prove that $$x = \frac{a}{3}$$.

\[ [\textbf{Hint :}\ \text{We have}\ \left( \dfrac{a + x}{a - x} \right)^{4} = 16 = 2^{4} \Rightarrow \dfrac{a + x}{a - x} = 2.\,] \]

Answer 1

Given: $$16\left(\frac{a - x}{a + x}\right)^3 = \left(\frac{a + x}{a - x}\right)$$

To prove: $$x = \frac{a}{3}$$

Proof:

1. $$16 = \left(\frac{a + x}{a - x}\right) \div \left(\frac{a - x}{a + x}\right)^3$$
2. or, $$16 = \left(\frac{a + x}{a - x}\right) \times \left(\frac{a + x}{a - x}\right)^3$$
3. or, $$\left(\frac{a + x}{a - x}\right)^4 = 16$$
4. or, $$\left(\frac{a + x}{a - x}\right)^4 = 2^4$$
5. or, $$\frac{a + x}{a - x} = 2$$ [Taking fourth root on both sides]
6. or, $$\frac{a + x}{a - x} = \frac{2}{1}$$
7. or, $$\frac{(a + x) + (a - x)}{(a + x) - (a - x)} = \frac{2 + 1}{2 - 1}$$ [By componendo and dividendo]
8. or, $$\frac{2a}{2x} = \frac{3}{1}$$
9. or, $$\frac{a}{x} = 3$$
10. or, $$3x = a$$
11. or, $$x = \frac{a}{3}$$

Hence proved.