Year 12 Methods (Unit 3 & 4) Exponential and logarithmic functions

Graphing logarithmic functions

Q: What are the features of the graph of y = log_a x?

For a base a greater than 1, the graph of y = log_a x has a vertical asymptote at x = 0 (the y-axis), crosses the x-axis at (1, 0), is always increasing, has domain x greater than 0 and range all real y. It is the mirror image of the exponential y = a to the power x in the line y = x.

Q: How does y = log_a(x - h) + k transform the graph?

The h shifts the graph sideways, so the vertical asymptote moves from x = 0 to x = h and the domain becomes x greater than h. The k shifts the graph up or down. The x-intercept is found where the argument equals 1, that is where x minus h equals 1.

Q: How are y = log_a x and y = a^x related?

They are inverse functions, so their graphs are reflections of each other in the line y = x. This swaps every coordinate pair: if (p, q) lies on y = a to the power x then (q, p) lies on y = log_a x. In particular y = ln x is the inverse of y = e to the power x.

Q: How do you solve a logarithmic equation like log_2 x = 5?

Rewrite it in index form: log_a x = c means x = a to the power c. So log_2 x = 5 gives x = 2 to the power 5 = 32, and ln x = 2 gives x = e squared, about 7.39. Equations such as ln x = 3 minus x are solved graphically by reading where the two graphs cross.

Q: What do reflections do to a logarithmic graph?

The graph of y = minus log_a x is y = log_a x reflected in the x-axis, so it decreases instead of increasing while keeping the same domain x greater than 0. The graph of y = log_a(minus x) is a reflection in the y-axis, which flips the domain to x less than 0.

Q: Where are logarithmic graphs used in the real world?

Logarithmic scales compress quantities that span many orders of magnitude. The Richter magnitude of an earthquake is R = log base 10 of (I over I0), the loudness of sound in decibels is a base-10 logarithm of intensity, and the pH of a solution is minus log base 10 of the hydrogen-ion concentration.

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In Year 12 Mathematical Methods (Queensland, QCAA), the graph of a logarithmic function \(y=\log_{a}x\) (with base \(a>1\), including \(y=\ln x\)) has a vertical asymptote \(x=0\), crosses the \(x\)-axis at \((1,0)\), is always increasing, with domain \(x>0\) and range all real \(y\) — studied here with its transformations \(y=\log_{a}(x-h)+k\), reflections, and its inverse relationship with \(y=a^{x}\) (a reflection in \(y=x\)).

A logarithmic function is \(y=\log_{a}x\), the inverse of the exponential \(y=a^{x}\) (base \(a>1\)); the natural log \(y=\ln x=\log_{e}x\) uses base \(e\approx 2.718\). Because a logarithm is only defined for a positive argument, the domain is \(x>0\) and the graph never crosses the \(y\)-axis — instead it plunges downward beside it, so \(x=0\) is a vertical asymptote.

Since \(\log_{a}1=0\), the graph always cuts the \(x\)-axis at the \(x\)-intercept \((1,0)\). With base \(a>1\) the curve is increasing (rising slowly to the right), and its range is all real \(y\). It is the mirror image of \(y=a^{x}\) in the line \(y=x\).

Transformations move this curve. For \(y=\log_{a}(x-h)+k\) the \(h\) slides it sideways, so the vertical asymptote moves to \(x=h\) and the domain becomes \(x>h\); the \(k\) slides it up or down. Reflections flip it: \(y=-\log_{a}x\) is a reflection in the \(x\)-axis (now decreasing), and \(y=\log_{a}(-x)\) is a reflection in the \(y\)-axis (domain \(x<0\)).

Key idea. \(y=\log_{a}x\) (\(a>1\)): vertical asymptote \(x=0\), \(x\)-intercept \((1,0)\), increasing, domain \(x>0\), range all real \(y\).

\(y=\log_{2}x\): increasing, through \((1,0)\), asymptote \(x=0\)

\(y=\log_{2}(x-2)\): shifted right \(2\), asymptote \(x=2\), \(x\)-intercept \((3,0)\)

The logarithmic function (base \(a>1\)), the inverse of \(y=a^{x}\):

\[y=\log_{a}x \qquad \log_{a}x=y \iff a^{y}=x \qquad (x>0)\]

y = \log_{a} x

The transformed logarithm (horizontal shift \(h\), vertical shift \(k\)):

\[y=\log_{a}(x-h)+k \qquad \text{vertical asymptote } x=h,\quad \text{domain } x>h\]

y = \log_{a} (x - h) + k

The inverse relationship with the exponential (reflection in \(y=x\)):

\[y=\log_{a}x \text{ is the inverse of } y=a^{x} \qquad y=\ln x \text{ is the inverse of } y=e^{x}\]

y = \ln x

Solving in index form. \(\log_{a}x=c\) gives \(x=a^{c}\). For example \(\log_{2}x=5\) gives \(x=2^{5}=32\), and \(\ln x=2\) gives \(x=e^{2}\approx 7.39\).

How to sketch \(y=\log_{a}(x-h)+k\)

Asymptote first. The graph exists only where the argument is positive, so \(x-h>0\); the vertical asymptote is \(x=h\) and the domain is \(x>h\).
Find the \(x\)-intercept. Set \(y=0\); with \(k=0\) this is where the argument equals \(1\), i.e. \(x-h=1\). (If \(k\neq0\), solve \(\log_{a}(x-h)=-k\).)
Find the \(y\)-intercept. Substitute \(x=0\) — but only if \(x=0\) lies in the domain \(x>h\).
Set the shape. With base \(a>1\) the curve increases (unless a negative coefficient reflects it in the \(x\)-axis), rising from the asymptote through the intercepts.

Reading reflections. A minus in front (\(y=-\log_{a}x\)) reflects in the \(x\)-axis, so the curve decreases; a minus inside (\(y=\log_{a}(-x)\)) reflects in the \(y\)-axis, so the domain becomes \(x<0\).

Example 1 — Features of \(y=\log_{2}x\)

State the vertical asymptote, the \(x\)-intercept, the domain, and whether \(y=\log_{2}x\) is increasing or decreasing.

Solution

Domain: \(\log_{2}x\) needs \(x>0\), so the vertical asymptote is \(x=0\).

\(x\)-intercept: \(\log_{2}x=0\Rightarrow x=1\), so \((1,0)\).

The base \(2>1\), so the graph is increasing; range all real \(y\).

\log_{2} 1 = 0

Example 2 — Transformation

For \(y=\ln(x-2)\), state the vertical asymptote, the domain and the \(x\)-intercept.

Solution

This is \(y=\ln x\) shifted right \(2\).

\(\text{asymptote}\)	\(:\)	\(x=2\)
\(\text{domain}\)	\(:\)	\(x>2\)
\(\ln(x-2)=0\)	\(\Rightarrow\)	\(x-2=1,\ x=3\)

Asymptote \(x=2\), domain \(x>2\), \(x\)-intercept \((3,0)\).

\ln (x - 2)

Example 3 — Solve a log equation

Solve for \(x\): (a) \(\log_{2}x=5\); (b) \(\ln x=2\). Give exact and \(2\) d.p. answers where needed.

Solution

(a) Rewrite in index form \(x=a^{c}\).

\(\log_{2}x\)	\(=\)	\(5\)
\(x\)	\(=\)	\(2^{5}=32\)

(b) Apply \(e\) to both sides (the inverse of \(\ln\)).

\(\ln x\)	\(=\)	\(2\)
\(x\)	\(=\)	\(e^{2}\approx 7.39\)

x = 2^{5} = 32

Example 4 — Richter-scale model

The Richter magnitude is \(R=\log_{10}\!\left(\dfrac{I}{I_{0}}\right)\). Find \(R\) when \(I=1000\,I_{0}\), and compare a magnitude-\(5\) with a magnitude-\(3\) quake.

Solution

\(R\)	\(=\)	\(\log_{10}1000=3\)
\(\dfrac{I_{5}}{I_{3}}\)	\(=\)	\(\dfrac{10^{5}I_{0}}{10^{3}I_{0}}=10^{2}\)

So \(R=3\), and a magnitude-\(5\) earthquake is \(100\) times as intense as a magnitude-\(3\) one.

\log_{10} 1000 = 3

Common pitfalls

The asymptote is vertical (\(x=h\)), not horizontal. Exponentials have a horizontal asymptote \(y=k\); logarithms have a vertical asymptote \(x=h\). For \(y=\log_{a}(x-h)+k\) it is \(x=h\), and the domain is \(x>h\).

\(y=\log_{a}x\) passes through \((1,0)\), not \((0,1)\). Because \(\log_{a}1=0\), the \(x\)-intercept is \((1,0)\). Swapping to \((0,1)\) is the exponential's \(y\)-intercept — the two graphs are inverses.

Check the domain before finding a \(y\)-intercept. A log graph only has a \(y\)-intercept if \(x=0\) is inside the domain. For \(y=\log_{2}(x-2)\) (domain \(x>2\)) there is no \(y\)-intercept.

Frequently asked questions

What are the features of the graph of y = log_a x?

For \(a>1\): a vertical asymptote \(x=0\), \(x\)-intercept \((1,0)\), always increasing, domain \(x>0\), range all real \(y\).

How does y = log_a(x - h) + k transform the graph?

\(h\) shifts it sideways (asymptote moves to \(x=h\), domain \(x>h\)) and \(k\) shifts it up or down. The \(x\)-intercept is where the argument equals \(1\).

How are y = log_a x and y = a^x related?

They are inverses, so their graphs are reflections in the line \(y=x\); every point \((p,q)\) on \(y=a^{x}\) becomes \((q,p)\) on \(y=\log_{a}x\). In particular \(y=\ln x\) is the inverse of \(y=e^{x}\).

How do you solve a logarithmic equation like log_2 x = 5?

Rewrite in index form: \(\log_{a}x=c\Rightarrow x=a^{c}\). So \(\log_{2}x=5\Rightarrow x=32\), and \(\ln x=2\Rightarrow x=e^{2}\approx 7.39\).

What do reflections do to a logarithmic graph?

\(y=-\log_{a}x\) reflects in the \(x\)-axis (now decreasing, same domain \(x>0\)); \(y=\log_{a}(-x)\) reflects in the \(y\)-axis (domain \(x<0\)).

Where are logarithmic graphs used in the real world?

On logarithmic scales such as the Richter scale \(R=\log_{10}(I/I_{0})\), sound loudness in decibels, and pH \(=-\log_{10}[\text{H}^{+}]\).

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