Sketch the graph of $y = 2\ln x$. In order to solve the equation $x e^{2x} = \dfrac{1}{3x}$, a suitable straight line has to be drawn on the same set of axes as the graph of $y = 2\ln x$. Find the equation of the straight line and the number of solution(s). Explain why the value of $x$ is between 0 and 1 for which $x^{2}e^{2x} = \dfrac{1}{3}$.

1. Identify and sketch the main features of $y=2\ln x$.

For $y=2\ln x$:

• Domain: $x>0$.
• As $x\to 0^+$, $\ln x\to -\infty$, so the $y$-axis is a vertical asymptote.
• When $x=1$, $y=2\ln 1=0$, so the graph passes through $(1,0)$.
• The graph is increasing for $x>0$.

Sketch an increasing logarithmic curve approaching the $y$-axis from the right and crossing the $x$-axis at $x=1$.

1. Take logarithms and express the given equation using $2\ln x$.

Given
$xe^{2x}=\frac{1}{3x}.$
Taking natural logarithms on both sides,
$\ln x+\ln(e^{2x})=\ln\left(\frac{1}{3x}\right).$
So
$\ln x+2x=-\ln 3-\ln x.$

2. Rearrange into the form of a straight line to be drawn with $y=2\ln x$.

Rearranging,
$2\ln x=-2x-\ln 3.$
Since the existing curve is
$y=2\ln x,$
the required straight line is
$y=-2x-\ln 3.$

3. Determine the number of intersections between the curve and the straight line.

The curve $y=2\ln x$ is increasing for $x>0$, while the line
$y=-2x-\ln 3$
is decreasing.

Also, as $x\to 0^+$, $2\ln x\to -\infty$, so the curve is below the line. At $x=1$,
$2\ln 1=0,$
while
$-2(1)-\ln 3<0,$
so the curve is above the line.

Hence the curve and the line intersect exactly once.

1. Relate the given equation to the graph intersection found in part (ii).

For $x>0$, taking logarithms of
$x^{2}e^{2x}=\frac13$
gives
$2\ln x+2x=-\ln 3,$
so
$2\ln x=-2x-\ln 3.$
Hence the solution is the intersection of $y=2\ln x$ and $y=-2x-\ln 3$.

2. Use the graph to justify that the intersection has $0<x<1$.

The curve $y=2\ln x$ exists only for $x>0$. From the sketch, the intersection with the straight line lies before the point $(1,0)$ on the curve, so its $x$-coordinate is less than $1$.

Therefore,
$0<x<1.$