If $a , b , c$ are real numbers $> 1$, then show that $$\frac { 1 } { 1 + \log _ { a ^ { 2 } b } \frac { c } { a } } + \frac { 1 } { 1 + \log _ { b ^ { 2 } c } \frac { a } { b } } + \frac { 1 } { 1 + \log _ { c ^ { 2 } a } \frac { b } { c } } = 3$$

Let $g$ be the function defined by
$$\begin{aligned} g : ] - \pi ; \pi [ & \longrightarrow \mathbf { C } \\ \theta & \longmapsto e ^ { \mathrm { i } x \theta } \int _ { 0 } ^ { + \infty } \frac { t ^ { x - 1 } } { 1 + t e ^ { \mathrm { i } \theta } } \mathrm {~d} t \end{aligned}$$
where $x$ is a fixed element of $]0;1[$. Deduce that
$$\int _ { 0 } ^ { + \infty } \frac { t ^ { x - 1 } } { 1 + t } \mathrm {~d} t = \frac { \pi } { \sin ( \pi x ) }$$

Recall that $x$ is a fixed element of $]0;1[$. Show that:
$$\int _ { 0 } ^ { + \infty } \frac { t ^ { x - 1 } } { 1 + t } \mathrm {~d} t = \int _ { 0 } ^ { 1 } \left( \frac { t ^ { x - 1 } } { 1 + t } + \frac { t ^ { - x } } { 1 + t } \right) \mathrm { d } t$$

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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The four real numbers $a , b , c$, and $d$ are all greater than 1 .
Suppose that they satisfy the equation $\log _ { c } d = \left( \log _ { a } b \right) ^ { 2 }$.
Use some of the lines given to construct a proof that, in this case, it follows that
$$( * ) \log _ { b } d = \left( \log _ { a } b \right) \left( \log _ { a } c \right)$$
(1) Let $x = \log _ { a } b$ and $y = \log _ { a } c$
(2) $d = \left( c ^ { x } \right) ^ { 2 }$
(3) $d = c ^ { \left( x ^ { 2 } \right) }$
(4) $d = b ^ { x y }$
(5) $d = \left( a ^ { y } \right) ^ { \left( x ^ { 2 } \right) }$
(6) $d = \left( \left( a ^ { y } \right) ^ { x } \right) ^ { 2 }$
(7) $d = \left( a ^ { x } \right) ^ { x y }$
(8) $d = a ^ { \left( y ^ { 2 x } \right) }$
(9) $d = a ^ { \left( x ^ { 2 } y \right) }$

Let t be a real number. The equalities
$$\begin{aligned} & x = e ^ { 2 \cos t } \\ & y = e ^ { 3 \sin t } \end{aligned}$$
are given.
Accordingly, which of the following gives the relationship between $x$ and y that is satisfied for every real number t?
A) $\ln ^ { 2 } x + \ln ^ { 2 } y = 4$
B) $\ln ^ { 2 } x + \ln ^ { 2 } y = 9$
C) $9 \ln ^ { 2 } x + 2 \ln ^ { 2 } y = 27$
D) $\ln ^ { 2 } x + 4 \ln ^ { 2 } y = 28$
E) $9 \ln ^ { 2 } x + 4 \ln ^ { 2 } y = 36$