Exponential Graph Intersections | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

An intersection gives equal y-values

(9709/22/O/N/24 Q4(a–b)) connects this chapter to modulus. Sketch $y=1+e^{2x}$ and $y=|x-4|$ using their intercepts and shapes.

At P, use |x − 4| = 4 − x before isolating e²ˣ.

The shown intersection $P$ lies to the left of 4. There, $x-4<0$ , so $|x-4|=4-x$ . Equal outputs give $1+e^{2x}=4-x$ .

Your turnPractice

For $y=2+e^{-x}$ , find the vertical-axis intercept and horizontal asymptote. Is the curve increasing or decreasing?

Show worked answer

It decreases through (0,3) and approaches $y=2$ from above as $x$ increases.

Isolate the exponential before taking its log

Worked example

Show that the intersection satisfies x = ½ ln(3 − x).

\begin{gathered}1+e^{2x}=4-x,\\e^{2x}=3-x,\\2x=\ln(3-x),\\x=\tfrac12\ln(3-x).\end{gathered}

The input $3-x$ must be positive, giving $x<3$ . Do not write $\ln3-\ln x$ : the subtraction is inside one log.

This is a rearrangement, not an explicit answer for $x$ , because $x$ still appears on both sides. Numerical solution methods come later.

Your turnPractice

Rearrange $2+3e^{2x}=8-x$ into $x=\dots$ with one logarithm on the right. State its input restriction.

Show worked answer

\begin{gathered}3e^{2x}=6-x\\e^{2x}=\frac{6-x}{3}\end{gathered}

Answer

\begin{gathered}x=\frac12\ln\left(\frac{6-x}{3}\right)\\x<6\end{gathered}

An intersection gives equal y-values

(9709/22/O/N/24 Q4(a–b)) connects this chapter to modulus. Sketch $y=1+e^{2x}$ and $y=|x-4|$ using their intercepts and shapes.

At P, use |x − 4| = 4 − x before isolating e²ˣ.

The shown intersection $P$ lies to the left of 4. There, $x-4<0$ , so $|x-4|=4-x$ . Equal outputs give $1+e^{2x}=4-x$ .

Your turnPractice

For $y=2+e^{-x}$ , find the vertical-axis intercept and horizontal asymptote. Is the curve increasing or decreasing?

Show worked answer

It decreases through (0,3) and approaches $y=2$ from above as $x$ increases.

Isolate the exponential before taking its log

Worked example

Show that the intersection satisfies x = ½ ln(3 − x).

\begin{gathered}1+e^{2x}=4-x,\\e^{2x}=3-x,\\2x=\ln(3-x),\\x=\tfrac12\ln(3-x).\end{gathered}

The input $3-x$ must be positive, giving $x<3$ . Do not write $\ln3-\ln x$ : the subtraction is inside one log.

This is a rearrangement, not an explicit answer for $x$ , because $x$ still appears on both sides. Numerical solution methods come later.

Your turnPractice

Rearrange $2+3e^{2x}=8-x$ into $x=\dots$ with one logarithm on the right. State its input restriction.

Show worked answer

\begin{gathered}3e^{2x}=6-x\\e^{2x}=\frac{6-x}{3}\end{gathered}

Answer

\begin{gathered}x=\frac12\ln\left(\frac{6-x}{3}\right)\\x<6\end{gathered}