Show that $y=\ln\left(\dfrac{e^{2}}{3x}\right)$ can be written in the form $y=a+b\ln(cx)$, where $a$, $b$ and $c$ are integers to be found. Hence, state a sequence of transformations which transform the graph of $y=\ln x$ onto the graph of $y=\ln\left(\dfrac{e^{2}}{3x}\right)$. The curve $y=f(x)$ passes through the point $P$ with coordinates $(a,b)$, where $b\ne 0$. The tangent to the curve at $P$ has gradient $5$. When $y=f(x)$ is transformed onto the curve $y=g(x)$, $P$ corresponds to the point $R$ on $y=g(x)$. For each of the following curves, state the coordinates of $R$ and find the gradient of the curve at $R$. $g(x)=2f(x-1)$ $g(x)=\dfrac{1}{f(x)}$

1. Express the given logarithmic function in the required form $y=a+b\ln(cx)$

$y = \ln\left(\frac{e^2}{3x}\right)$
$y = \ln(e^2) - \ln(3x)$
$y = 2 - \ln(3x)$
Comparing with $y = a + b\ln(cx)$, we have $a = 2$, $b = -1$ and $c = 3$.

2. State the sequence of transformations mapping $y=\ln x$ to $y=2-\ln(3x)$.

To transform $y = \ln x$ to $y = 2 - \ln(3x)$:

1. Replace $x$ with $3x$: A scaling parallel to the $x$-axis by a scale factor of $\frac{1}{3}$ yields $y = \ln(3x)$.
2. Multiply the function by $-1$: A reflection in the $x$-axis yields $y = -\ln(3x)$.
3. Add $2$ to the function: A translation of $2$ units in the positive $y$-direction yields $y = 2 - \ln(3x)$.

1. Apply the transformations to find the coordinates of $R$.

The transformation $x o x-1$ translates the graph $1$ unit to the right, mapping the $x$-coordinate $a$ to $a+1$.
Multiplying the function by $2$ stretches the graph vertically by a factor of $2$, mapping the $y$-coordinate $b$ to $2b$.
Thus, the coordinates of $R$ are $(a+1, 2b)$.

2. Differentiate $g(x)$ to relate its gradient to $f'(x)$.

$g'(x) = \frac{d}{dx}[2f(x-1)] = 2f'(x-1)$

3. Evaluate the gradient of $g(x)$ at the point $R$.

At point $R$, the $x$-coordinate is $a+1$.
Substituting $x = a+1$ into the derivative:
$g'(a+1) = 2f'(a+1-1) = 2f'(a)$
Given that the gradient of $f(x)$ at $P$ is $5$, we have $f'(a) = 5$.
Gradient at $R = 2(5) = 10$.

1. Determine the coordinates of the corresponding point $R$ on the new curve.

For $y = g(x) = \frac{1}{f(x)}$, the $x$-coordinate remains unchanged.
Substitute $x = a$ to find the new $y$-coordinate:
$g(a) = \frac{1}{f(a)}$
Since $P(a, b)$ lies on $f(x)$, $f(a) = b$.
$g(a) = \frac{1}{b}$
The coordinates of $R$ are $\left(a, \frac{1}{b}\right)$.

2. Find the gradient of the transformed curve at $R$.

To find the gradient, differentiate $g(x)$ with respect to $x$ using the chain rule:
$g'(x) = \frac{d}{dx}\left([f(x)]^{-1}\right) = -[f(x)]^{-2}f'(x) = -\frac{f'(x)}{[f(x)]^2}$
We are given the gradient of $f(x)$ at $P$ is $5$, so $f'(a) = 5$.
Substitute $x = a$, $f(a) = b$, and $f'(a) = 5$ into the derivative:
$g'(a) = -\frac{5}{b^2}$