Exponents and logarithms: study notes

1. Numbers of the form $a \times 10^k$

A number is in standard form when it is written as $a \times 10^k$ with $1 \le a < 10$ and $k$ an integer. Multiply or divide such numbers by dealing with the two parts separately: $(3 \times 10^5)(4 \times 10^{-2}) = 12 \times 10^{3} = 1.2 \times 10^4$, re-adjusting so that $a$ stays in $[1, 10)$. To add or subtract, first write both numbers with the same power of ten: $5.1 \times 10^6 + 3 \times 10^5 = 5.1 \times 10^6 + 0.3 \times 10^6 = 5.4 \times 10^6$. A calculator will display results in this form; copy the exponent correctly and round $a$ to the accuracy asked for (three significant figures unless told otherwise). None of this is in the booklet; it is expected as fluent arithmetic.

2. Laws of exponents (learn these, not in the booklet)

For a base $a > 0$ and any rational exponents $m, n$: $a^m a^n = a^{m+n}$, $\dfrac{a^m}{a^n} = a^{m-n}$, $(a^m)^n = a^{mn}$, $(ab)^n = a^n b^n$, $a^0 = 1$, $a^{-n} = \dfrac{1}{a^n}$, $a^{1/n} = \sqrt[n]{a}$ and $a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$. For example $8^{-2/3} = \dfrac{1}{(\sqrt[3]{8})^2} = \dfrac{1}{4}$ and $27^{4/3} = 3^4 = 81$. The most useful exam skill is rewriting different bases as powers of one base: $4^x = 2^{2x}$, $9^{x+1} = 3^{2x+2}$, $\sqrt{3} = 3^{1/2}$, $\dfrac{1}{25} = 5^{-2}$. Once every term is a power of the same base, an equation such as $2^{3x-1} = 4^{x+2}$ becomes $3x - 1 = 2x + 4$, so $x = 5$, with no logarithms needed.

3. What a logarithm is; base 10 and base $e$

The booklet gives the defining relationship $a^x = b \iff x = \log_a b$ for $a > 0$, $b > 0$, $a \ne 1$. In words, $\log_a b$ is the exponent to which $a$ must be raised to give $b$: $\log_2 32 = 5$ because $2^5 = 32$, and $\log_5 \dfrac{1}{25} = -2$. Two consequences you must learn (not in the booklet): $\log_a 1 = 0$ and $\log_a a = 1$; the general identities $\log_a a^k = k$ and $a^{\log_a b} = b$ are in the booklet, but under Topic 2 (SL 2.9), so know them anyway. Base 10 is written $\log x$ and base $e$ is written $\ln x$; both are calculator keys and are used for numerical evaluation, for example $\ln 7 \approx 1.95$ and $\log 250 \approx 2.40$. The domain condition is essential and the booklet only hints at it through the condition $b > 0$ beside the definition: $\log_a x$ exists only for $x > 0$. Questions state domains such as $x > 2$ so that laws may be applied, and markschemes usually award a mark for rejecting a root that makes any argument zero or negative. The graph of $y = \log_a x$ for $a > 1$ passes through $(1, 0)$, increases slowly, and has the $y$-axis as a vertical asymptote; it is the reflection of $y = a^x$ in the line $y = x$.

4. Laws of logarithms and change of base (in the booklet)

For positive $x, y$ and valid base $a$: $\log_a xy = \log_a x + \log_a y$, $\log_a \dfrac{x}{y} = \log_a x - \log_a y$, $\log_a x^m = m \log_a x$, and change of base $\log_b a = \dfrac{\log_c a}{\log_c b}$. Learn how to use them in both directions. Combining: $2\log_3 x - \log_3 (x+1) = \log_3 \dfrac{x^2}{x+1}$. Splitting: $\ln \dfrac{\sqrt{x}}{e^2} = \dfrac{1}{2}\ln x - 2$. A constant can be written as a log in any base, $3 = \log_2 8$, which is often the key step when one side of an equation is a number. Change of base lets you evaluate $\log_3 20 = \dfrac{\ln 20}{\ln 3}$ on a calculator, and it gives the reciprocal result $\log_b a = \dfrac{1}{\log_a b}$ (take $c = a$). It is also how mixed-base equations are solved: $\log_4 x = \dfrac{\log_2 x}{\log_2 4} = \dfrac{1}{2}\log_2 x$. A standard symbolic task gives $\log_a 2 = p$ and $\log_a 3 = q$ and asks for $\log_a 18 = p + 2q$ or $\log_a \dfrac{3}{\sqrt{2}} = q - \dfrac{1}{2}p$; break the argument into prime factors first. There is no law for $\log_a (x + y)$.

5. Solving exponential equations

If both sides can be written as powers of the same base, equate exponents. Otherwise take logarithms of both sides (any base, usually $\ln$) and bring the exponent down with the power law: $5^{x} = 12 \Rightarrow x \ln 5 = \ln 12 \Rightarrow x = \dfrac{\ln 12}{\ln 5} \approx 1.54$, which is the same as $x = \log_5 12$. When the unknown appears in two different bases, collect the $x$ terms: $3^{x+1} = 2^{2x}$ gives $(x + 1)\ln 3 = 2x \ln 2$, so $x(2\ln 2 - \ln 3) = \ln 3$ and $x = \dfrac{\ln 3}{\ln 4 - \ln 3}$. In exact-answer questions leave the result as a quotient of logs or as a single log; a decimal with no working typically earns nothing. The same technique drives the 'smallest $n$ such that' questions in sequences and finance: $2000(1.04)^n > 5000$ becomes $n \ln 1.04 > \ln 2.5$, so $n > 23.4$ and the smallest integer is $n = 24$. Remember that dividing by $\ln r$ with $0 < r < 1$ reverses the inequality because $\ln r < 0$. On calculator papers you may also solve numerically with a graph or table, but write the inequality down first for the method mark.

6. Hidden quadratics and log equations with a domain check

An equation such as $3^{2x} - 10 \cdot 3^x + 9 = 0$ is a quadratic in disguise: write $3^{2x} = (3^x)^2$, let $u = 3^x$, solve $u^2 - 10u + 9 = 0$ to get $u = 1$ or $u = 9$, then return to $x$: $3^x = 1 \Rightarrow x = 0$ and $3^x = 9 \Rightarrow x = 2$. A value $u \le 0$ must be rejected because $a^x > 0$ for all real $x$. Watch for disguises: $4^x = (2^x)^2$, $e^{2x} = (e^x)^2$, $2^{x+1} = 2 \cdot 2^x$ and $9^x - 3^{x+1} = (3^x)^2 - 3 \cdot 3^x$. Similarly $(\log_2 x)^2 - 3\log_2 x + 2 = 0$ with $u = \log_2 x$ gives $u = 1$ or $u = 2$, so $x = 2$ or $x = 4$; note that $(\log x)^2$ is not $\log x^2$. For equations containing several logs, use the laws to reach a single log equal to a constant or to another single log, remove the logs with the definition, solve the resulting polynomial, and finally check every root against the original domain, writing a reason such as 'reject $x = -2$ since $\log_2 x$ is undefined'. That written reason is normally a separate mark.

Formulas

Worked examples

Can you...

• Convert between $a^x = b$ and $x = \log_a b$ instantly, evaluate logs such as $\log_3 81$ and $\ln e^{-2}$ without a calculator, and state that $\log_a 1 = 0$ and $\log_a a = 1$?
• Write $4^x$, $9^{x+1}$, $\sqrt{3}$ and $\dfrac{1}{25}$ as single powers of $2$, $3$, $3$ and $5$, and use this to spot a hidden quadratic in $a^x$?
• Combine $2\log_3 x - \log_3(x+1)$ into one log, and split $\ln \dfrac{\sqrt{x}}{e^2}$ into separate terms, quoting the booklet laws correctly?
• Solve $\log_2 x + \log_2(x - 2) = 3$ completely, including rejecting the invalid root with a written reason?
• Take logs of both sides of $5^x = 12$ or $3^{x+1} = 2^{2x}$, solve for $x$ exactly as a quotient of logs, and turn $A(1.04)^n > B$ into the smallest integer $n$?