6.3.3 Using Log Graphs in Modelling

Logarithmic and Exponential Graphs

The curve starts steep and becomes less steep as $x$ increases, but it never becomes horizontal.
Asymptote at $x = 0$ .

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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 $x = -3$ .

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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.

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(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:

\frac{1}{3 - \log_2{3}}.

Solution:

Given:

\begin{align*} y &= 2 \times 3^x \quad \text{(Equation 1)} \\ y &= 8^x \quad \text{(Equation 2)} \end{align*}

Set the two equations equal to find the intersection point:

2 \times 3^x = 8^x

Take the logarithm of both sides (base 2):

\log_2{(2 \times 3^x)} = \log_2{(8^x)}

Apply the logarithm rules:

\log_2{2} + \log_2{3^x} = x \times \log_2{8}

Simplify using the power rule:

1 + x \log_2{3} = 3x \quad \text{(since } \log_2{8} = 3\text{)}

Rearrange to solve for $x$ :

1 = x(3 - \log_2{3})

Thus,

x = \frac{1}{3 - \log_2{3}}

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