How is e^x related to other exponential bases like 10^x?

Any positive base can be written using e. Since 10 = e to the power (ln 10), it follows that 10 to the power x = e to the power (x ln 10). So y = 10 to the power x is y = e to the power x dilated horizontally by a factor of 1 divided by ln 10 from the y-axis.

How is e^x related to other exponential bases like 10^x?

Any positive base can be written using e. Since 10 = e to the power (ln 10), it follows that 10 to the power x = e to the power (x ln 10). So y = 10 to the power x is y = e to the power x dilated horizontally by a factor of 1 divided by ln 10 from the y-axis.

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

The exponential functionf(x)=ex.

Q: What is the number e?

e is an irrational constant approximately equal to 2.718. It is the unique base a for which the gradient expression (a to the power h, minus 1) divided by h approaches 1 as h approaches 0, which makes y = e to the power x the natural exponential function used throughout calculus.

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

The graph of y = e to the power x passes through (0, 1), has the horizontal asymptote y = 0 (the x-axis), is always increasing, has domain all real x and range y greater than 0. Because the base e is greater than 1 the curve rises to the right and flattens towards the x-axis to the left.

Q: How does a transformation change the graph of y = e^x?

For y = e to the power (x minus h) plus k, the value h shifts the graph sideways and k shifts it vertically, moving the horizontal asymptote to y = k so the range becomes y greater than k. For y = A e to the power (kx) plus c, A dilates it vertically (and reflects it in the x-axis if A is negative) and c sets the asymptote y = c.

Q: How do you solve an equation such as e^x = 12?

Use the natural logarithm, which is the inverse of e to the power x. Taking ln of both sides of e to the power x = 12 gives x = ln 12, the exact answer, which is approximately 2.48 by technology.

Q: When does N = N0 e^(kt) model growth or decay?

In N = N0 e to the power (kt) the sign of k decides. When k is greater than 0 the quantity grows exponentially; when k is less than 0 it decays towards 0; when k equals 0 it is constant. N0 is the initial amount at time t = 0.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), the natural exponential function \(y=e^x\) uses the constant \(e\approx 2.718\). Its graph passes through \((0,1)\), has the horizontal asymptote \(y=0\), is always increasing with range \(y>0\) — studied here with its transformations \(y=e^{x-h}+k\), solving \(e^x=k\) using \(\ln\), and growth-and-decay models \(N=N_0e^{kt}\), ready for the calculus of Unit 3.

The natural exponential function is \(y=e^x\), where the base is the constant \(e\approx 2.718\). The number \(e\) is the special base for which the gradient property \(\dfrac{e^{h}-1}{h}\to 1\) as \(h\to 0\); this is what makes \(e^x\) the natural choice for the calculus that follows. Because \(e^0=1\), the graph passes through the \(y\)-intercept \((0,1)\).

Since the base \(e>1\), the curve is always increasing: it races upward to the right and flattens towards the \(x\)-axis to the left without ever reaching it — the line \(y=0\) is a horizontal asymptote. The domain is all real \(x\) and the range is \(y>0\).

Transformations shift and stretch this curve. For \(y=e^{x-h}+k\) the asymptote moves to \(y=k\); for \(y=A\,e^{kx}+c\) the constant \(c\) sets the asymptote and \(A\) dilates (and reflects, if \(A<0\)). Because any positive base can be written using \(e\) (for example \(10^x=e^{x\ln 10}\)), \(e^x\) underlies every exponential model.

Key idea. \(y=e^x\) (\(e\approx 2.718\)): \(y\)-intercept \((0,1)\), asymptote \(y=0\), always increasing, domain all real \(x\), range \(y>0\).

\(y=e^x\): increasing, through \((0,1)\), asymptote \(y=0\)

\(y=e^x-2\): shifted down \(2\), asymptote \(y=-2\), \(y\)-intercept \((0,-1)\)

The natural exponential function (base \(e\approx 2.718\)):

\[y=e^x \qquad e=\lim_{n\to\infty}\left(1+\tfrac1n\right)^n\approx 2.718\]

y = e^{x}

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

\[y=e^{x-h}+k \qquad \text{asymptote } y=k\]

y = e^{x - h} + k

The growth-and-decay model over time \(t\), with initial amount \(N_0\):

\[N=N_0\,e^{kt} \qquad k>0:\ \text{growth} \quad k<0:\ \text{decay}\]

N = N_{0} e^{k t}

Solving with \(\ln\). The natural logarithm is the inverse of \(e^x\), so \(e^x=k\) gives \(x=\ln k\). For example \(e^x=12\) gives \(x=\ln 12\approx 2.48\).

How to sketch \(y=e^{x-h}+k\)

Asymptote first. The \(+k\) lifts the whole curve, so the horizontal asymptote is \(y=k\) (not \(y=0\)).
Find the \(y\)-intercept. Substitute \(x=0\) into \(y=e^{x-h}+k\) and evaluate with technology if needed.
Set the shape. With base \(e>1\) the curve increases (unless reflected by a negative coefficient); the range is \(y>k\).

Relating to other bases. Any positive base can be written with \(e\): since \(10=e^{\ln 10}\), \(10^x=e^{x\ln 10}\). So \(y=10^x\) is \(y=e^x\) dilated horizontally by \(\dfrac{1}{\ln 10}\) from the \(y\)-axis.

Example 1 — Features of \(y=e^x\)

State the \(y\)-intercept, horizontal asymptote, range, and whether \(y=e^x\) is increasing or decreasing.

Solution

\(y\)-intercept: at \(x=0\), \(e^0=1\), so \((0,1)\).

Asymptote: \(y=0\); range: \(y>0\).

The base \(e\approx 2.718>1\), so the graph is increasing.

e^{0} = 1

Example 2 — Transformation

For \(y=e^{-x}+2\), state the horizontal asymptote, the \(y\)-intercept and the range.

Solution

This is \(y=e^x\) reflected in the \(y\)-axis then shifted up \(2\).

\(\text{asymptote}\)	\(:\)	\(y=2\)
\(y(0)\)	\(=\)	\(e^{0}+2=3\)
\(\text{range}\)	\(:\)	\(y>2\)

Asymptote \(y=2\), \(y\)-intercept \((0,3)\), range \(y>2\) (decreasing).

y = e^{0} + 2 = 3

Example 3 — Solve with \(\ln\)

Solve for \(x\): (a) \(e^x=12\); (b) \(2e^x=10\). Give exact and \(2\) d.p. answers.

Solution

(a) Take \(\ln\) of both sides.

\(e^x\)	\(=\)	\(12\)
\(x\)	\(=\)	\(\ln 12\approx 2.48\)

(b) Divide by \(2\) first, then take \(\ln\).

\(e^x\)	\(=\)	\(5\)
\(x\)	\(=\)	\(\ln 5\approx 1.61\)

x = \ln 12

Example 4 — Decay model with base \(e\)

A sample has mass \(M=80\,e^{-0.1t}\) grams after \(t\) days. Find the initial mass and the mass after \(10\) days.

Solution

\(M(0)\)	\(=\)	\(80\,e^{0}=80\)
\(M(10)\)	\(=\)	\(80\,e^{-1}\)
	\(\approx\)	\(29.43\)

Initial mass \(80\) g; after \(10\) days about \(29.43\) g. Since \(k=-0.1<0\) this is decay.

80 e^{- 1} \approx 29.43

Common pitfalls

\(y=e^x\) passes through \((0,1)\), not \((0,e)\). Because \(e^0=1\), the \(y\)-intercept is \((0,1)\). The value \(e\approx 2.718\) is \(y\) when \(x=1\), i.e. the point \((1,e)\).

The asymptote moves with \(k\), not \(h\). For \(y=e^{x-h}+k\) the horizontal asymptote is \(y=k\), so the range is \(y>k\). The \(h\) only slides the graph sideways.

Growth vs decay is the sign of the exponent. In \(N=N_0e^{kt}\), \(k>0\) grows and \(k<0\) decays. A model like \(e^{-0.1t}\) decays because the exponent is negative.

Frequently asked questions

What is the number e?

\(e\approx 2.718\) is an irrational constant — the unique base \(a\) for which \(\dfrac{a^{h}-1}{h}\to 1\) as \(h\to 0\). This makes \(y=e^x\) the natural exponential function.

What are the features of the graph of y = e^x?

It passes through \((0,1)\), has asymptote \(y=0\), is always increasing, with domain all real \(x\) and range \(y>0\).

How does a transformation change the graph of y = e^x?

For \(y=e^{x-h}+k\), \(h\) shifts it sideways and \(k\) vertically, moving the asymptote to \(y=k\). For \(y=A\,e^{kx}+c\), \(c\) sets the asymptote and \(A\) dilates (reflecting if \(A<0\)).

How do you solve an equation such as e^x = 12?

Take \(\ln\) of both sides (the inverse of \(e^x\)): \(e^x=12\) gives \(x=\ln 12\approx 2.48\).

When does N = N0 e^(kt) model growth or decay?

\(k>0\) gives growth, \(k<0\) gives decay, \(k=0\) is constant. \(N_0\) is the amount at \(t=0\).

How is e^x related to other bases like 10^x?

Since \(10=e^{\ln 10}\), \(10^x=e^{x\ln 10}\). So \(y=10^x\) is \(y=e^x\) dilated horizontally by \(\dfrac{1}{\ln 10}\) from the \(y\)-axis.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

Start practising

← Previous subtopic

Revision of exponential functions