Define Euler's number e as the base for which y=e^x has gradient 1 at the y-intercept, work with the natural logarithm ln x = log_e x as the inverse of e^x, sketch y=e^x and y=ln x as reflections in y=x, transform y=e^x, and solve e^x=k and ln x=k by converting between the two forms

What this dot point is asking

The earlier pages built every exponential $y = a^x$ and every logarithm $y = \log_a x$ for a general base. This page singles out one special base, the number $e \approx 2.718$, and the logarithm that goes with it, the natural logarithm $\ln x$. NESA expects you to know what makes $e$ special (the curve $y = e^x$ crosses the $y$-axis with a gradient of exactly $1$), to treat $\ln x = \log_e x$ as the inverse of $e^x$ and so convert freely between the two, to sketch $y = e^x$ and $y = \ln x$ as reflections in the line $y = x$, to transform $y = e^x$ by the usual shifts and reflections, and to solve equations of the form $e^x = k$ and $\ln x = k$. The pay-off is huge: $e$ and $\ln$ are the natural language of growth and decay, and almost every real growth or decay model in Year 12, from population to radioactive decay to cooling, is written with $e$. Getting comfortable with "apply $\ln$ to undo $e^x$, apply $e^x$ to undo $\ln$" now makes all of that routine later.

One thing this page deliberately does not do is differentiate $e^x$. The gradient-$1$-at-the-$y$-intercept property is stated as a definition of $e$, seen on the graph as a tangent, not derived with calculus. The rule $\frac{d}{dx}e^x = e^x$ and the rest of the calculus belong to the introduction to differentiation module; here $e$ and $\ln$ are functions and graphs.

The answer

Why one base is special: the gradient at the y-intercept

Every exponential $y = a^x$ shares the same skeleton from the previous page: it passes through $(0, 1)$, has the $x$-axis as a horizontal asymptote, and increases for a base $a > 1$. But the curves differ in steepness, and in particular they differ in how steeply they cross the $y$-axis. Imagine drawing the tangent line to $y = a^x$ right at the point $(0, 1)$ and measuring its gradient. That gradient depends on the base:

• For $y = 2^x$, the tangent at $(0, 1)$ has gradient about $0.69$, so the curve crosses the $y$-axis fairly gently.
• For $y = 3^x$, the tangent at $(0, 1)$ has gradient about $1.10$, so it crosses more steeply.

So a base of $2$ gives a gradient under $1$ and a base of $3$ gives a gradient over $1$. Somewhere between $2$ and $3$ there must be a base for which the tangent at $(0, 1)$ has gradient exactly $1$. That base is the number we call $e$, and to four decimal places $e \approx 2.7183$. It is an irrational number, like $\pi$, with a never-ending non-repeating decimal; for the exam $e \approx 2.718$ is plenty.

The function $y = e^x$ is so important it gets a name of its own, the exponential function (with "the", to set it apart from all the other exponential functions $y = a^x$). Its defining feature is that single clean tangent: gradient $1$ at the point $(0, 1)$.

The graph of y = e^x and its tangent

Since $e \approx 2.718$ lies between $2$ and $3$, the graph of $y = e^x$ has exactly the standard increasing-exponential shape, sitting between $y = 2^x$ and $y = 3^x$. Its features are the familiar ones:

• The $y$-intercept is $(0, 1)$, because $e^0 = 1$.
• At $x = 1$ the height is $e$, the point $(1, e) \approx (1, 2.72)$, because $e^1 = e$.
• The $x$-axis $y = 0$ is a horizontal asymptote: as $x \to -\infty$, $e^x \to 0$ from above. As $x \to \infty$ the curve rises ever more steeply.
• The domain is all real $x$ and the range is $y > 0$.

What sets this curve apart visually is the tangent at $(0, 1)$. Because its gradient is $1$, that tangent is the line $y = x + 1$: starting from $(0, 1)$ it rises one unit up for every one unit across, passing through $(-1, 0)$. The figure shows $y = e^x$ with this tangent drawn in; notice how the line "kisses" the curve at $(0, 1)$ and matches its steepness exactly there.

A neat consequence, worth knowing as a feature of this graph rather than proving: for $y = e^x$ the gradient at any point equals the height of the curve there. At $(0, 1)$ the height is $1$ and the gradient is $1$; that is the special case you can see, and it holds all along the curve. The reason (that $e^x$ is its own derivative) is the calculus that comes later.

The natural logarithm ln x as the inverse of e^x

Just as $\log_2 x$ undoes $2^x$, the logarithm to base $e$ undoes $e^x$. Logarithms to base $e$ are so common they get a special name and symbol: the natural logarithm, written $\ln x$. So

Read $\ln x$ as "the power of $e$ that gives $x$". Because $e$ and $\ln$ are inverse operations, applying one straight after the other cancels:

These two cancellation rules are the engine of the whole topic. The first says that taking $\ln$ of $e^x$ leaves just the index $x$; the second says raising $e$ to the power $\ln x$ gives back $x$. A few values follow immediately from "the power of $e$ that gives this number":

• $\ln 1 = 0$, because $e^0 = 1$.
• $\ln e = 1$, because $e^1 = e$.
• $\ln(e^2) = 2$, because $e^2$ is $e$ to the power $2$.

For a number that is not a neat power of $e$, such as $\ln 5$, you use the calculator's $\ln$ key; it is a genuine number, $\ln 5 = 1.609438\ldots$, just as $\sqrt 2$ is.

Why y = e^x and y = ln x are reflections in y = x

Because $e^x$ and $\ln x$ are inverse functions, their graphs are exact mirror images in the line $y = x$, exactly as for $2^x$ and $\log_2 x$ on the previous page. Reflecting in $y = x$ swaps each point $(p, q)$ to $(q, p)$, and that swap is the index-and-log relationship: a point $(p, q)$ on $y = e^x$ says $e^p = q$, and its mirror $(q, p)$ on $y = \ln x$ says $\ln q = p$, the very same statement read backwards. So the table of values for the two curves is the same with the columns swapped:

The reflection also swaps the key features. The exponential's horizontal asymptote $y = 0$ becomes the logarithm's vertical asymptote $x = 0$; its $y$-intercept $(0, 1)$ becomes the logarithm's $x$-intercept $(1, 0)$; and its domain (all real $x$) and range ($y > 0$) swap to give the logarithm domain $x > 0$ and range all real $y$. There is even a tangent bonus: reflecting the tangent $y = x + 1$ (gradient $1$ at $(0, 1)$ on $e^x$) gives the line $y = x - 1$, the tangent of gradient $1$ at $(1, 0)$ on $\ln x$. The diagram draws both curves and the mirror line.

Converting between e^x and ln to solve equations

The whole reason $\ln$ is useful is that it turns an exponential equation into a one-line solution, and $e$ does the reverse for a log equation. The pattern is always "apply the inverse to both sides":

• To solve $e^x = k$ (with $k > 0$), take $\ln$ of both sides: $x = \ln k$. This is exact; press the $\ln$ key for a decimal.
• To solve $\ln x = k$, raise $e$ to both sides: $x = e^k$. Again exact, with the $e^x$ key for a decimal.

For example, $e^x = 5$ gives $x = \ln 5 = 1.609438\ldots \approx 1.61$, and $\ln x = 2$ gives $x = e^2 = 7.389056\ldots \approx 7.39$. If the exponential is not alone, isolate it first: $3e^x = 21$ becomes $e^x = 7$, then $x = \ln 7 = 1.945910\ldots \approx 1.95$. The same applies to logs: $2\ln x = 6$ becomes $\ln x = 3$, then $x = e^3 = 20.085537\ldots \approx 20.09$. Always leave the answer in exact form ($\ln k$ or $e^k$) unless the question asks for a decimal, because the exact form is what "find the exact value" rewards.

Transforming the graph of y = e^x

The number $e$ is just a particular base, so every transformation from the previous page applies to $y = e^x$ unchanged; the trick is the same, track the asymptote, the intercept and the range as you move the curve.

• Translation up or down by $k$: $y = e^x + k$ shifts the curve $k$ units up (down if $k$ is negative). The horizontal asymptote moves from $y = 0$ to $y = k$, and the range becomes $y > k$. So $y = e^x + 2$ has asymptote $y = 2$, intercept $(0, 3)$ and range $y > 2$.
• Translation left or right by $h$: $y = e^{x - h}$ shifts $h$ units right ($h$ negative shifts left). The asymptote stays $y = 0$ and the range stays $y > 0$.
• Reflection in the $y$-axis: $y = e^{-x}$ flips the curve left to right, turning growth into decay. The $y$-intercept stays $(0, 1)$, the asymptote stays $y = 0$ and the range stays $y > 0$; only the direction reverses, so now the curve falls from left to right and hugs the axis on the right. At $x = 1$, $y = e^{-1} \approx 0.37$, and at $x = -1$, $y = e^{1} \approx 2.72$. This is the natural shape of decay, used for cooling and radioactive decay.
• Reflection in the $x$-axis: $y = -e^x$ flips the curve below the axis; the asymptote stays $y = 0$ but the range becomes $y < 0$.

The most-tested transformation is the vertical shift, because it is the one that moves the asymptote, and a sketch that does not relabel the asymptote loses the mark.

How exam questions ask about e and natural logarithms

The command words map onto specific actions:

• "Find the exact value of $x$ if $e^x = k$" wants $x = \ln k$ left in that form, not a rounded decimal. "Exact" is the signal to stop at $\ln k$ or $e^k$.
• "Solve $\ln x = k$" wants $x = e^k$, again exact unless a decimal is requested; if it says "to two decimal places", convert at the end.
• "Sketch $y = e^x$" means draw the increasing curve, label $(0, 1)$ and ideally $(1, e)$, and draw the asymptote $y = 0$ as a dashed line with its equation; a tangent of gradient $1$ at $(0, 1)$ shows you know what makes $e$ special.
• "On the same axes, sketch $y = e^x$ and $y = \ln x$" is asking you to use the reflection in $y = x$: draw the exponential, draw the dashed line $y = x$, reflect to get the logarithm, and mark a mirror pair such as $(0, 1)$ and $(1, 0)$.
• "Describe the transformation that maps $y = e^x$ to $\ldots$" wants the named move (translate up/down/left/right, reflect in an axis) and the amount, with the new asymptote stated.
• A growth or decay context ("$P = P_0 e^{kt}$", "find when ...") is solved by isolating the exponential and taking $\ln$; state the answer with its units and round sensibly.

This page builds directly on exponential and logarithmic graphs and on logarithms and the laws of logarithms; if the reflection in $y = x$ or the cancellation rules feel slippery, re-reading "a logarithm is the index" is the surest fix. The calculus of $e^x$ then follows in differentiating exponential functions.