Modelling with Exponentials & Logarithms (Edexcel A-Level Mathematics): Revision Notes

Example: Sketching the graph of $y = \ln(x + 3)$

• Start by recognizing the base graph: $y = \ln(x)$.
• To sketch $y = \ln(x + 3)$, perform a translation of $(-3, 0)$ on the base graph. The graph will shift 3 units to the left, moving the asymptote to $x = -3$.

Example: Sketch the graph of $y = 2e^{3x}$

• Always a good idea to have a "start graph" then apply transformations to it.
• Start with the graph of $y = e^x$ (in red).
• Apply the following transformations:
• Stretch by scale factor 2, parallel to the $y$-axis: This gives us the graph of $y = 2e^x$ (in blue).
• Stretch by scale factor $\frac{1}{3}$ parallel to the $x$-axis: This gives us the graph of $y = 2e^{3x}$ (in green).
• The final graph is $y = 2e^{3x}$ (shown in green on the right-hand side).
• Key feature: The asymptote of the graph is at $y = 0$.

Q3. (OCR 4722, Jun 2008, Q8)

(i) Sketch the curve $y = 2 \times 3^x$, stating the coordinates of any intersections with the axes.

Sketch:

• The curve $y = 2 \times 3^x$ intersects the $y$-axis at $(0, 2)$.
• There is no intersection with the $x$-axis as the curve approaches zero asymptotically but never actually touches the $x$-axis. (ii) The curve $y = 2 \times 3^x$ intersects the curve $y = 8^x$ at the point $P$. Show that the $x$-coordinate of $P$ may be written as:

Solution:

Given:

Set the two equations equal to find the intersection point:

Take the logarithm of both sides (base 2):

Apply the logarithm rules:

Simplify using the power rule:

Rearrange to solve for $x$:

Thus,

This is the required $x$-coordinate of the intersection point $P$.