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## This Week

In lectures, we started chapter 3 and are in section 3.2. We also looked at some simpler proofs of Theorems 1.3.2 and 3.1.4. These are reproduced below.

In tutorials we did p.36 Q 6, p.42 Q.2, 4 and  p.44 Q.1 (a)(e), 2(a).

## Sample Test

Hopefully by Tuesday morning.

Here we give an alternative proof of Theorem 3.1.4. It is very likely that you will be asked to prove some of Theorem 3.1.4 in the exam and you are free to choose which proof you prefer. You may even prove parts (ii) and (iii) of Theorem 3.1.4 using the proof in the notes and then use this proof for part (iv). In class I also gave an easier proof of Theorem 1.3.2 although this won’t be examinable.

### Theorem 3.1.4

For all positive $a$ and $b$, and any rational number $r$, we have

1. $\ln 1=0$,
2. $\ln(ab)=\ln a+\ln b$,
3. $\displaystyle\ln\left(\frac{a}{b}\right)=\ln a-\ln b$,
4. $\ln a^r=r\ln a$.

Proof : The proof of 1. is trivial.

2. Consider the functions $y_1(x)=\ln ax$ and $y_2(x)=\ln x$.

Note that $\displaystyle y'_1(x)=\frac{1}{ax}\times a=1/x$ and $y_2'(x)=1/x$. Therefore $y_1$ and $y_2$ have the same slope so by Theorem 1.3.2 they only differ by a constant, say $C$ $y_1(x)=y_2(x)+C$ for all $x>0$.

Now let $x=1$: $\ln (a)=\ln(1)+C\Rightarrow \ln(a)=0+C\Rightarrow C=\ln a$. $\Rightarrow \ln (ax)=\ln(x)+\ln (a)$ for all $x>0$.

Now let $x=b$; $\ln(ab)=\ln(a)+\ln(b)$.

3. Now by 2. we have that $\ln\left(\frac{a}{b}\right)=\ln\left(a\cdot\frac1{b}\right)=\ln a+\ln\left(\frac{1}{b}\right)$.  (*)

However $\ln \left(b\cdot\frac1{b}\right)=\ln(b)+\ln\left(\frac{1}{b}\right)$.

But $\ln(b/b)=\ln 1=0$ by 1. Hence $\ln b+\ln\left(\frac1{b}\right)=0\Rightarrow \ln\left(\frac{1}{b}\right)=-\ln (b)$.

Putting this back into (*): $\ln \left(\frac{a}{b}\right)=\ln a-\ln b$.

4. Consider the functions $y_1(x)=\ln (x^r)$ and $y_2(x)=r\ln x$.

Note that $\displaystyle y_1'(x)=\frac{1}{x^r}\times (rx^{r-1})=\frac{r}{x^r}$ and $y_2'(x)=\frac{r}{x^r}$. Hence these functions have the same slope and thus differ only by a constant, say $C$: $y_1(x)=y_2(x)+C$ for all $x>0$.

Let $x=1$; $\ln(1^r)=r\ln(1)+C\Rightarrow \ln 1=0+C\Rightarrow C=0$.

Thence $y_1(x)=y_2(x)$ for all $x>0$ and particular for any $a>0$: $\ln (a^r)=r\ln a$ $\bullet$