# Multiplying reciprocal fractions with exponents

The question is as follows:

If we multiply a fraction with its reciprocal then we get 1. So by this logic this should be equal to 1^(a/a+b).

But, the solutions I have been provided with states the answer to be

Where am I going wrong?

Thanks

• You've multiplied before applying the exponent. – Randall Oct 14 '18 at 3:22
• Can you calculate $2\cdot3^2$? – zipirovich Oct 14 '18 at 3:40

In your work, you've computed something like $$(st)^u$$ when the traditional order of operations would have had you compute $$s(t^u)$$. This is why you got $$1$$ (incorrectly).
What they want you to do is make use of the fact that negating the exponent causes the fraction to flip. So, you were to take the second factor and rewrite as $$\left(\frac{y}{x}\right)^{a/(a+b)} = \left(\frac{x}{y}\right)^{-a/(a+b)}.$$ Now multiply by $$\frac{x}{y}$$ to get $$\frac{x}{y}\left(\frac{y}{x}\right)^{a/(a+b)} = \frac{x}{y}\left(\frac{x}{y}\right)^{-a/(a+b)} = \left(\frac{x}{y}\right)^{1-\frac{a}{a+b}}.$$
Now that exponent is $$1 - \frac{a}{a+b} = \frac{a+b}{a+b} - \frac{a}{a+b} = \frac{a+b-a}{a+b} = \frac{b}{a+b}$$. Hence the final answer is $$\left(\frac{x}{y}\right)^{\frac{b}{a+b}}$$ as claimed.