- The variables \(x\) and \(y\) are connected by the relation \(y = ax^n\), where \(a\) and \(n\) are constants; \(y = 3\) when \(x = 4\) and \(y = 2\) when \(x = 9\). Find the exact values of \(n\) and \(a\).
Dividing the second by the first to eliminate \(a\) gives \[\frac{2}{3} = \left(\frac{9}{4}\right)^n.\]
To work with this, it will be more convenient to write everything as powers of \(2\) and \(3\).
That is, \(2^{2n+1}=3^{2n+1}\), which can only be the case when \(2n+1=0\) (as \(2\) and \(3\) have highest common factor \(1\)).
Therefore, \(n=-\frac{1}{2}\).
Now we can go back to equation \(\eqref{eq:1}\) or equation \(\eqref{eq:2}\) to find that \(a=6\).
Alternatively, if you know about logarithms…
Taking logs to base \(10\) with our starting equations (any base will do, in fact) we get \[\log3 = \log a + 2n\log2, \log2 = \log a + 2n \log3.\]
Subtracting these gives \(\log3 - \log2 = 2n(\log2-\log3)\), and so \(n = -\dfrac{1}{2}\).
Substituting back into one of our starting equations gives \(a = 6\).
- Express \[\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\] as a fraction having a rational denominator.
If we multiply both the numerator and the denominator by \(\sqrt{a} + \sqrt{b}\), then the fraction becomes \[\dfrac{(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{b})}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}=\frac{(\sqrt{a}+\sqrt{b})^2}{a-b}=\dfrac{a+b+2\sqrt{ab}}{a-b}.\] This has a rational denominator.
- Show that, if \(n\) is a positive integer, then \(n(n+2)\) lies between \(n^2\) and \((n+1)^2\).
Note \(n(n+2)-n^2 = 2n > 0\), so \(n(n+2)>n^2\).
We also have \[n(n+2)-(n+1)^2=n^2 + 2n - n^2 - 2n -1=-1 < 0,\] so \(n(n+2)<(n+1)^2\).
So \(n(n+2)\) lies between \(n^2\) and \((n+1)^2\).
- If \(n\) is a positive integer, use the results in (i) and (ii) to find in terms of \(n\) two consecutive integers between which \[\dfrac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}-\sqrt{n}}\] must lie.
By using (i) we get \[\dfrac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}-\sqrt{n}}=\dfrac{2n+2+2\sqrt{n(n+2)}}{2}=n+1+\sqrt{(n+2)n}.\]
Now we know \((n+1)^2>n(n+2)>n^2\), which implies \(n+1>\sqrt{n(n+2)}>n\).
So adding \(n+1\) to each part of the inequality, we get \[2n+2>n+1+\sqrt{(n+2)n}>2n+1,\]
and so \(\dfrac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}-\sqrt{n}}\) lies between \(2n+1\) and \(2n+2\).
