2.2, 3.2 - The functions exp and ln | Cambridge A-Level Mathematics

The natural exponential and natural logarithm undo one another. In this shared part of 2.2 and 3.2, you will connect their graphs, solve equations with unknown exponents and handle exponential inequalities, including decreasing powers. The sign of a logarithm will explain when an inequality reverses.

Natural exponential growth

Can an exponential curve approach zero without ever reaching it? Powers such as $2^{-10}=1/1024$ suggest how: increasingly negative exponents produce increasingly small positive numbers. The natural exponential has the same essential behaviour, using a special base called $e$ .

Natural exponential

The number $e$ is a mathematical constant, approximately $2.71828$ . The natural exponential function is $\exp x=e^x$ . Its domain is all real numbers and its range is $y>0$ .

Like every positive-base power, $e^0=1$ , $e^{-x}=1/e^x$ and $e^{u+v}=e^ue^v$ . Because $e>1$ , $e^x$ increases as $x$ increases. Negative exponents give positive values below $1$ , never negative values.

For $y=e^{kx}$ , the sign of $k$ determines the direction: positive $k$ makes the graph increase; negative $k$ makes it decrease. Both pass through $(0,1)$ . When $k=0$ , the rule is simply $y=1$ .

Worked example 1: an increasing exponential

Question. Describe a sketch of $y=e^{2x}$ , identifying its intercept and horizontal asymptote.

Plan. Use three exact values and examine increasingly negative inputs.

x=-1,0,1\quad\Longrightarrow\quad y=e^{-2},1,e^2.

These values are approximately $0.135,1,7.39$ . As $x$ decreases without bound, $e^{2x}$ tends to zero while remaining positive.

Answer. An increasing curve through $(0,1)$ , with horizontal asymptote $y=0$ to the left and no $x$ -intercept.

Check. Increasing $x$ by $1$ multiplies the output by $e^2>1$ , confirming the direction for every input.

Change only the sign of $k$ . This exchanges the left and right behaviour.

Worked example 2: a decreasing exponential

Question. Describe the same features of $y=e^{-2x}$ .

Plan. Use the same inputs and recognise the reciprocal of the previous outputs.

x=-1,0,1\quad\Longrightarrow\quad y=e^2,1,e^{-2}.

As $x$ increases without bound, $e^{-2x}=1/e^{2x}$ tends to zero from above.

Answer. A decreasing curve through $(0,1)$ , with horizontal asymptote $y=0$ to the right and no $x$ -intercept.

Check. Each unit increase of $x$ multiplies the output by $e^{-2}<1$ , so the positive outputs decrease. The graph is the reflection of $y=e^{2x}$ in the $y$ -axis.

The curves share a range and intercept, but approach zero in opposite directions.

[DIAGRAM: asset_name: Growth and decay with the same intercept; asset_slug: exponential-growth-decay; caption: Changing the sign in the exponent reverses the direction while every output stays positive; description: Two equal-scale panels show y equals e to the power 2x and y equals e to the power minus 2x. Both pass through 0 comma 1. Exact points at x equals minus 1 and 1 are labelled using e squared and e to the power minus 2. The horizontal axis is the asymptote y equals 0.]
Changing the sign in the exponent reverses the direction while every output stays positive

Watch out

$e^{-2x}$ is $1/e^{2x}$ , not $-e^{2x}$ . At $x=0$ , the first is $1$ , while the second would be $-1$ .

Use the sign of the coefficient and one exact input.

Natural logarithms

Every positive number is the output of $e^x$ for exactly one real input. Reversing that function gives the natural logarithm, written $\ln$ . It answers “what power of $e$ gives this positive number?”

Natural logarithm

The natural logarithm is the logarithm to base $e$ :

\ln u=v\quad\Longleftrightarrow\quad e^v=u,\qquad u>0.

The graph of $y=\ln x$ is the reflection of $y=e^x$ in $y=x$ . Its domain is $x>0$ and its range is all real numbers.

Reflection swaps $(0,1)$ with $(1,0)$ . It also swaps the exponential's horizontal asymptote $y=0$ with the logarithm's vertical asymptote $x=0$ . The logarithm increases: it is negative for $0<x<1$ , zero at $1$ , and positive for $x>1$ . It has no $y$ -intercept.

[DIAGRAM: asset_name: An exponential and its inverse; asset_slug: exponential-log-inverse; caption: Reflection in y equals x swaps exponential inputs and outputs; description: Equal-scale axes show y equals e to the x in blue, y equals ln x in sage, and a dashed mirror line y equals x. The points 0 comma 1 and 1 comma 0, and 1 comma e and e comma 1 are marked as reflected pairs. The exponential approaches y equals 0 on the left; the logarithm approaches x equals 0 from the right.]
Reflection in y equals x swaps exponential inputs and outputs

The inverse relationship supplies two cancellation rules.

Natural inverse identities

\ln(e^v)=v\quad(v\in\mathbb R),\qquad e^{\ln u}=u\quad(u>0).

The usual logarithm laws also apply to $\ln$ . These identities are not listed in the formula booklet: learn them.

Choose the inverse operation according to the outside function.

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2.2, 3.2 - The functions exp and ln

Natural exponential growth

Natural logarithms

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