1.5 Exponential and Logarithmic Functions

Big idea

A polynomial grows because its input is raised to a fixed power. An exponential function grows because the input is the power. That single swap changes the behavior completely: a polynomial adds a roughly constant amount over equal steps once the leading term dominates, while an exponential multiplies by a constant factor over equal steps. Constant percentage change is the signature, and it is why populations, interest, drug concentration and radioactive decay are all modelled this way.

Because $b^x$ is increasing and never repeats a value when $b > 1$, it has an inverse, and that inverse is the logarithm. A logarithm answers the question “what exponent do I need”, which is exactly what an equation with the unknown upstairs is asking. Everything hard about logarithms becomes easy once you keep that sentence in front of you: the logarithm is the exponent.

One base gets special treatment. Among all the choices of $b$, exactly one makes the graph of $b^x$ cross the $y$-axis with slope $1$, and that number is $e$, approximately $2.718$. Choosing it makes the derivative of $e^x$ equal to $e^x$, the cleanest derivative in the subject, so $e$ and the natural logarithm $\ln$ become the default forms for the rest of the course even when a problem is stated with another base.

This section also supplies the algebra you will need under a derivative sign. The laws of logarithms turn products into sums, quotients into differences, and powers into coefficients. Later that means an intimidating product or quotient can be flattened before differentiating, and an exponential equation can be brought down to a linear one. The manipulations are worth practising now, because in the next chapter they appear as one line in the middle of a longer problem.

Decoder

For a base $b > 0$ with $b \ne 1$, the statement $\log_b x = y$ means exactly the same thing as $b^y = x$.

The two equations are one fact written in two arrangements, and translating between them solves most log problems on sight. Asked for $\log_2 32$, read it as “$2$ to what power is $32$” and answer $5$. Asked to solve $\log_3 x = 4$, translate to $x = 3^4 = 81$.

The translation also explains the domains for free. The left equation accepts only $x > 0$, because the right equation shows $x$ as a power of a positive base, and such powers are always positive and never zero. So $\log_b x$ is undefined at $0$ and at every negative number, its graph has a vertical asymptote at $x = 0$, and its range is all real numbers - the mirror image of the exponential, whose domain is all reals and range is the positive numbers.

The classic mistake is inventing laws by analogy. There is no rule that turns $\log(x + y)$ into anything simpler, and $\log(x)/\log(y)$ is not $\log(x - y)$. The real laws all come from the laws of exponents, and each one converts an operation on the inputs into a simpler operation on the outputs, never the other way round.

Definitions and results

Exponential functions. For $b > 0$ with $b \ne 1$, the function $f(x) = b^x$ has domain all real numbers, range $(0, \infty)$, and $y$-intercept $1$. It is increasing when $b > 1$ and decreasing when $0 < b < 1$, and the $x$-axis is a horizontal asymptote on one side.

Laws of exponents. For any base and any real exponents,

$$ b^{m}b^{n} = b^{m+n}, \qquad \frac{b^m}{b^n} = b^{m-n}, \qquad (b^m)^n = b^{mn}, \qquad b^{-n} = \frac{1}{b^n} $$

These are the source of every logarithm law below.

The number $e$. The value $e \approx 2.71828$ is the base for which $b^x$ has slope $1$ at $x = 0$. It also arises as the limit of $(1 + 1/n)^n$ as $n$ grows without bound, which is why continuously compounded interest is written $P e^{rt}$.

Logarithms as inverses. $\log_b x$ is the inverse of $b^x$, so $\log_b(b^x) = x$ for all real $x$ and $b^{\log_b x} = x$ for all $x > 0$. The natural logarithm $\ln x$ is $\log_e x$; the common logarithm $\log x$ is base $10$.

Laws of logarithms. For $M, N > 0$ and any real $p$,

$$ \log_b(MN) = \log_b M + \log_b N, \qquad \log_b\!\left(\frac{M}{N}\right) = \log_b M - \log_b N, \qquad \log_b(M^p) = p\log_b M $$

Also $\log_b 1 = 0$ and $\log_b b = 1$.

Change of base. Any logarithm converts to natural logarithms by $\log_b x = \ln x / \ln b$. This is how a calculator with only $\ln$ and $\log$ evaluates $\log_7 20$, and how base-$b$ problems get folded into the natural-log machinery.

Solving exponential equations. If both sides can be written as powers of one base, set the exponents equal. Otherwise take the logarithm of both sides and use the power law to bring the unknown exponent down as a coefficient.

Growth and decay models. A quantity changing at a constant percentage rate satisfies $A(t) = A_0 e^{kt}$, with $k > 0$ for growth and $k < 0$ for decay. A stated doubling time or half-life determines $k$ through one equation.

Worked examples

A transformed exponential graph

Describe $g(x) = 2^{x+1} - 3$.

Start from $y = 2^x$. The $+1$ inside shifts the graph left by $1$; the $-3$ outside lowers it by $3$, which drags the horizontal asymptote from $y = 0$ down to $y = -3$. The function is still increasing, so the range is $(-3, \infty)$ and the domain stays all real numbers.

The $y$-intercept is $g(0) = 2^1 - 3 = -1$. For the $x$-intercept, solve $2^{x+1} = 3$, so $x + 1 = \log_2 3$ and $x = \log_2 3 - 1 \approx 0.585$. That sits just right of the origin, consistent with a graph that passes below the axis at $x = 0$ and climbs.

Combining and separating logarithms

Evaluate $\log_2 80 - \log_2 5$. The quotient law collapses it: $\log_2(80/5) = \log_2 16 = 4$. Translating back confirms it, since $2^4 = 16$.

Now expand $\log\!\left(\dfrac{x^3\sqrt{y}}{z^2}\right)$ for positive $x, y, z$. The quotient law splits the fraction, the product law splits the numerator, and the power law pulls each exponent out front, with the square root counted as the exponent $1/2$:

$$ 3\log x + \tfrac{1}{2}\log y - 2\log z $$

Run it backwards as a check: the coefficients become exponents, the sum becomes a product, and the difference restores the denominator.

Two exponential equations

Solve $5 \cdot 3^{2x} = 45$. Divide first, so $3^{2x} = 9 = 3^2$. Same base, so $2x = 2$ and $x = 1$. Substituting back, $5 \cdot 3^2 = 45$.

Solve $4^{x+1} = 7$. No common base is available, so take natural logarithms of both sides and use the power law:

$$ (x+1)\ln 4 = \ln 7, \qquad x = \frac{\ln 7}{\ln 4} - 1 \approx 1.404 - 1 = 0.404 $$

Check numerically: $4^{1.404} \approx 7$, as required. Notice that the logarithm was not applied to each term separately; it was applied to both complete sides, which is the only legal move.

A doubling-time model

A culture starts at $200$ cells and doubles every $6$ hours. Model it as $A(t) = 200e^{kt}$ with $t$ in hours.

Doubling gives $A(6) = 400$, so $e^{6k} = 2$ and $k = (\ln 2)/6 \approx 0.1155$.

When does the count reach $500$? Solve $200e^{kt} = 500$, so $e^{kt} = 2.5$ and

$$ t = \frac{\ln 2.5}{k} = \frac{6\ln 2.5}{\ln 2} \approx \frac{6(0.9163)}{0.6931} \approx 7.93 \text{ hours} $$

Check the size of the answer: $500$ is between one doubling ($400$ at $6$ hours) and two doublings ($800$ at $12$ hours), and $7.93$ sits in that window, nearer the lower end as it should be.

Practice

Start with the shape of the graphs. Identify base, direction, asymptote, intercepts and range for exponential functions and their shifted versions.

Next, the algebra. Expand a single logarithm into pieces, condense pieces back into one, and change base when the problem asks for a number.

Last, equations with the unknown in the exponent. Try a common base first; take logarithms of both sides when no common base exists.

Quiz

Six items on exponential graphs and their features, the laws of logarithms, and solving exponential equations.