Exponential and Logarithmic Graphs

On the same system of axes, draw the graphs of \(f(x) = 10^{x}\) and its inverse \(f^{-1}(x) = \log{x}\). Investigate the properties of \(f\) and \(f^{-1}\).

Determine the properties of \(f(x)\)

• Function: \(y = 10^{x}\)
• Shape: increasing graph
• Intercept(s): \((0;1)\)
• Asymptote(s): horizontal asymptote at \(x\)-axis, line \(y = 0\)
• Domain: \(\{x: x \in \mathbb{R} \}\)
• Range: \(\{y: y > 0, y \in \mathbb{R} \}\)

Draw the graphs

The graph of the inverse \(f^{-1}\) is the reflection of \(f\) about the line \(y = x\).

Determine the properties of \(f^{-1}(x)\)

• Function: \(y = \log{x}\)
• Shape: increasing graph
• Intercept(s): \((1;0)\)
• Asymptote(s): vertical asymptote at \(y\)-axis, line \(x = 0\)
• Domain: \(\{x: x > 0, x \in \mathbb{R} \}\)
• Range: \(\{y: y \in \mathbb{R} \}\)

Notice that the inverse is a function: \(f^{-1}(x) = \log{x}\) is a one-to-one function since every input value is associated with only one output value.

1. Draw a sketch of \(g(x) = \log_{10}{x}\).
2. Reflect the graph of \(g\) about the \(x\)-axis to give the graph \(h\).
3. Investigate the properties of \(h\).
4. Use \(g\) and \(h\) to suggest a general conclusion.

Sketch the graph of \(g(x)= \log_{10}{x}\)

Reflect \(g\) about the \(x\)-axis

An easy method for reflecting a graph about a certain line is to imagine folding the Cartesian plane along that line and the reflected graph is pressed onto the plane.

Investigate the properties of \(h\)

• Function: passes the vertical line test
• Shape: decreasing graph
• Intercept(s): \((1;0)\)
• Asymptote(s): vertical asymptote at \(y\)-axis, line \(x = 0\)
• Domain: \(\{x: x > 0, x \in \mathbb{R} \}\)
• Range: \(\{y: y \in \mathbb{R} \}\)

Since \(h(x)\) is symmetrical to \(g(x)\) about the \(x\)-axis, this means that every \(y\)-value of \(g\) corresponds to a \(y\)-value of the opposite sign for \(h\).

\begin{align*} \text{Given } g(x) & = \log_{10}{x} \\ \therefore h(x) & = - \log_{10}{x} \\ \text{Let } y & = - \log_{10}{x} \\ -y & = \log_{10}{x} \\ \therefore 10^{-y} & = x \\ ( \cfrac{1}{10} )^{y} & = x \\ \therefore y & = \log_{\cfrac{1}{10}}{x} \\ \therefore h(x) & = - \log_{10}{x} = \log_{\cfrac{1}{10}}{x} \end{align*}

1. Draw a sketch of \(h(x) = \log_{\cfrac{1}{10}}{x}\).
2. Draw the graph of \(r(x)\), the reflection of \(h\) about the line \(y = x\).
3. Investigate the properties of \(r\).
4. Write down the new equation if \(h\) is shifted \(\text{1}\) unit upwards and \(\text{2}\) units to the right.

Sketch the graph of \(h(x)= \log_{\cfrac{1}{10}}{x}\)

Reflect \(h\) about the line \(y = x\)

Investigate the properties of \(r\)

• Function: passes the vertical line test
• Shape: decreasing graph
• Intercept(s): \((0;1)\)
• Asymptote(s): horizontal asymptote at \(x\)-axis, line \(y = 0\)
• Domain: \(\{x: x \in \mathbb{R} \}\)
• Range: \(\{y: y > 0, y \in \mathbb{R} \}\)

Since \(h(x)\) is symmetrical to \(r(x)\) about the line \(y = x\), this means that \(r\) is the inverse of \(h\).

\begin{align*} h(x) & = \log_{\cfrac{1}{10}}{x} \\ \text{Let } y & = \log_{\cfrac{1}{10}}{x} \\ \text{Inverse: } x & = \log_{\cfrac{1}{10}}{y} \\ \therefore ( \cfrac{1}{10} )^{x} & = y \\ {10}^{-x} & = y \\ & \\ \therefore r(x) = h^{-1}(x)& = {10}^{-x} \end{align*}

Therefore, \(r(x)\) is an exponential function of the form \(y = b^{x}\) with \(0 < b < 1\). In words, the base \(b\) is a positive fraction with a value between \(\text{0}\) and \(\text{1}\).

Vertical and horizontal shifts

If \(h\) is shifted \(\text{1}\) unit upwards and \(\text{2}\) units to the right, then the new equation will be: \[y = \log_{\cfrac{1}{10}}{(x - 2)} + 1\]

The vertical asymptote is \(x = 2\) and the horizontal asymptote is \(y = 1\).