\section*{2. For ALL APPLICANTS.}
In this question you may use without proof the following fact:

$$\ln ( 1 - x ) = - x - \frac { x ^ { 2 } } { 2 } - \frac { x ^ { 3 } } { 3 } - \frac { x ^ { 4 } } { 4 } \cdots - \frac { x ^ { n } } { n } \ldots \quad \text { for any } x \text { with } | x | < 1 .$$

[Note that $\ln x$ is alternative notation for $\log _ { e } x$.]\\
(i) By choosing a particular value of $x$ with $| x | < 1$, show that

$$\ln 2 = \frac { 1 } { 2 } + \frac { 1 } { 2 \times 2 ^ { 2 } } + \frac { 1 } { 3 \times 2 ^ { 3 } } + \frac { 1 } { 4 \times 2 ^ { 4 } } + \frac { 1 } { 5 \times 2 ^ { 5 } } + \ldots$$

(ii) Use part (i) and the fact that

$$\frac { 1 } { n 2 ^ { n } } < \frac { 1 } { 3 \times 2 ^ { n } } \quad \text { for } n \geqslant 4$$

to find the integer $k$ such that $\frac { k } { 24 } < \ln 2 < \frac { k + 1 } { 24 }$.\\
(iii) Show that

$$\ln \left( \frac { 3 } { 2 } \right) = \frac { 1 } { 2 } - \frac { 1 } { 2 \times 2 ^ { 2 } } + \frac { 1 } { 3 \times 2 ^ { 3 } } - \frac { 1 } { 4 \times 2 ^ { 4 } } + \frac { 1 } { 5 \times 2 ^ { 5 } } - \ldots$$

and deduce that

$$\ln 3 = 1 + \frac { 1 } { 3 \times 2 ^ { 2 } } + \frac { 1 } { 5 \times 2 ^ { 4 } } + \frac { 1 } { 7 \times 2 ^ { 6 } } + \ldots$$

(iv) Deduce that $\frac { 13 } { 12 } < \ln 3 < \frac { 11 } { 10 }$.\\
(v) Which is larger: $3 ^ { 17 }$ or $4 ^ { 13 }$ ? Without calculating either number, justify your answer.



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