Exponentials, logarithms and sequences

Growth by a constant factor

A quantity that doubles every $T$ units of time is multiplied by 2 once per period, and $\frac{t}{T}$ counts the periods. Starting from $N_0$,

$$N(t) = N_0 \cdot 2^{t/T}.$$

A half-life works the same way with a factor of $\frac{1}{2}$: $N(t) = N_0 \cdot \left(\frac{1}{2}\right)^{t/T}$. Any constant factor fits this pattern; growth of 5% per period multiplies by 1.05 each time, as in percentages and successive changes.

The usual question gives the start and the end and asks how long. Divide first, so that only the growth factor is left, then write it as a power. If active users double every 3 months, growing from 5,000 to 160,000 takes

$$\frac{160\,000}{5\,000} = 32 = 2^5 \quad\Longrightarrow\quad \frac{t}{3} = 5 \quad\Longrightarrow\quad t = 15 \text{ months}.$$

The exponent, 5, is the number of doublings; the ratio, 32, is not a time at all. Halving runs the other way: 400 mg of a substance with a half-life of 6 hours becomes $400 \cdot \left(\frac{1}{2}\right)^{24/6} = \frac{400}{16} = 25$ mg after 24 hours.

Solving by matching powers

Calculators are not allowed on the test, so you cannot take the logarithm of an awkward number. The hand method is to write both sides as powers of one base and compare exponents: if $a^x = a^y$ with $a > 0$ and $a \ne 1$, then $x = y$. The exponent rules do the rewriting:

Rule	Example
$a^m \cdot a^n = a^{m+n}$	$2^3 \cdot 2^4 = 2^7$
$(a^m)^n = a^{mn}$	$8^x = (2^3)^x = 2^{3x}$
$a^{-n} = \frac{1}{a^n}$	$\frac{1}{9} = 3^{-2}$
$a^{1/n} = \sqrt[n]{a}$	$27^{1/3} = 3$

So $9^{x+1} = 27^x$ becomes $3^{2x+2} = 3^{3x}$, giving $2x + 2 = 3x$ and $x = 2$. Check: $9^3 = 729 = 27^2$. A few powers known by heart make the rewriting quick: $2^{10} = 1024$, $3^4 = 81$, $3^5 = 243$, $5^4 = 625$.

Logarithms are exponents

$\log_a y = x$ means exactly $a^x = y$. So $\log_2 32 = 5$, $\log_{10} 1000 = 3$ and $\log_3 \frac{1}{9} = -2$. Because a log is an exponent, each exponent rule becomes a log rule:

• Product: $\log_a (xy) = \log_a x + \log_a y$.
• Quotient: $\log_a \frac{x}{y} = \log_a x - \log_a y$.
• Power: $\log_a (x^k) = k \log_a x$.
• Special values: $\log_a a = 1$ and $\log_a 1 = 0$.

The rules let awkward logs combine into whole numbers: $\log_6 4 + \log_6 9 = \log_6 36 = 2$, and $\log_2 48 - \log_2 3 = \log_2 16 = 4$. To change base, divide: $\log_a b = \frac{\log_c b}{\log_c a}$, so $\log_4 8 = \frac{\log_2 8}{\log_2 4} = \frac{3}{2}$. There is no rule for $\log_a (x + y)$: the log of a sum does not split.

Domain first, then solve, then check

$\log_a y$ exists only when $y > 0$. Combining logs can hide this, so write the domain down before you start. In

$$\log_2 x + \log_2 (x - 6) = 4,$$

both arguments must be positive, so $x > 6$. The product rule gives $\log_2 \big(x(x - 6)\big) = 4$, so $x^2 - 6x = 2^4 = 16$, and $(x - 8)(x + 2) = 0$. The combined log also exists at $x = -2$, where $x(x - 6) = 16$ is positive, but the original logs do not, so $-2$ is rejected. Only $x = 8$ remains: $\log_2 8 + \log_2 2 = 3 + 1 = 4$.

Arithmetic sequences: add the same step

In 7, 11, 15, 19, … each term adds the same difference, $d = 4$. Reaching the $n$-th term takes $n - 1$ steps, not $n$:

$$a_n = a_1 + (n - 1)d.$$

For the sum, pair the first term with the last, the second with the second-last, and so on. Every pair has the same total, and $n$ terms make $\frac{n}{2}$ pairs (when $n$ is odd, the middle term on its own is worth half a pair):

$$S_n = \frac{n}{2}(a_1 + a_n).$$

The integers from 1 to 100 add to $50 \times 101 = 5050$. To count the terms of a list such as 12, 15, …, 99, count the steps and add one: $\frac{99 - 12}{3} + 1 = 30$.

Geometric sequences: multiply by the same ratio

In 5, 15, 45, 135, … each term multiplies by the same ratio, $r = 3$. In general

$$a_n = a_1 r^{n-1}, \qquad S_n = a_1 \cdot \frac{1 - r^n}{1 - r} \quad (r \ne 1).$$

The sum formula comes from subtracting $rS_n$ from $S_n$: every middle term cancels, leaving $a_1 - a_1 r^n$. Check: $5 + 15 + 45 + 135 = 200$, and $5 \cdot \frac{1 - 81}{1 - 3} = 5 \cdot 40 = 200$.

When $|r| < 1$, the power $r^n$ shrinks towards 0, so the sum of infinitely many terms settles at

$$S_\infty = \frac{a_1}{1 - r}.$$

For $12 + 6 + 3 + 1.5 + \cdots$, $r = \frac{1}{2}$ and $S_\infty = \frac{12}{1 - \frac{1}{2}} = 24$. When $|r| \ge 1$ the terms never shrink and there is no finite sum. A geometric sequence is exponential growth recorded once per period: a quantity that doubles each period, measured at the end of each one, is a geometric sequence with $r = 2$.

Recursive rules

Some sequences give each term from the one before. When the rule mixes multiplying and adding, look for a constant that turns it into a pure geometric sequence. With $a_{n+1} = 3a_n - 2$ and $a_1 = 4$, the terms are 4, 10, 28, 82, …; subtracting 1 from each gives 3, 9, 27, 81, the powers of 3, so $a_n = 3^n + 1$. The differences of such a sequence (6, 18, 54) are geometric too, which is another way to spot it.