TMUA Logarithms & Exponentials: Laws & Tricks

Logarithms feel harder than they are because the notation hides a simple idea. A logarithm is nothing more than an index, written the other way round. The statement $\log_a b = c$ says exactly the same thing as $a^c = b$, so every fact about logs is really a fact about powers that you already know from indices. Once that link clicks, the whole topic collapses from a list of strange rules into one familiar idea seen from a new angle. This guide covers what the TMUA actually tests, the log laws with worked examples, how to change base, how to solve exponential equations cleanly, and the traps that quietly cost marks.

What the TMUA actually examines

Logarithms and exponentials sit in the AS-level pure content that dominates Paper 1, and they thread through several of the TMUA syllabus topics. You are expected to know the log laws, to change between bases, to solve equations where the unknown sits in an exponent, and to handle the natural logarithm $\ln x$ as the inverse of $e^x$. None of this is exotic, but the examiners test whether you can wield it fluently rather than recite it.

The questions rarely say "evaluate this logarithm" in isolation. Instead a log or an exponential is the lever that unlocks an otherwise stuck equation. You meet a population that doubles, a quantity that decays, an equation like $2^x = 7$ where $x$ is trapped upstairs, and the only way down is to take logs. On Paper 2 the same machinery appears inside reasoning and proof, where a single misapplied log law sinks an entire argument. The skill being tested is not memory, it is judgement: knowing which law turns the expression in front of you into something you can solve.

So, as with so much of the paper, the goal is not new knowledge. It is to make the familiar instant.

The log laws, with worked examples

Everything rests on three laws, and all three fall straight out of the index laws once you remember that a log is an index.

The product law says $\log_a(xy) = \log_a x + \log_a y$. Multiplying numbers adds their logs, exactly mirroring how $a^m \times a^n = a^{m+n}$ adds the indices. The quotient law is its partner: $\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y$. Dividing subtracts, just as $\frac{a^m}{a^n} = a^{m-n}$ subtracts. The power law is the workhorse: $\log_a(x^n) = n \log_a x$. A power inside the log comes out as a multiplier in front, and this is the rule that lets you free an unknown from an exponent.

Two facts round out the set. Since $a^1 = a$, we have $\log_a a = 1$, and since $a^0 = 1$, we have $\log_a 1 = 0$. These look trivial but they end arguments cleanly.

A worked example that uses several at once. Write $\log_2 24 - \log_2 3$ as a single number. The quotient law collapses the difference: $\log_2 24 - \log_2 3 = \log_2\left(\frac{24}{3}\right) = \log_2 8$. Now read that as an index question: $2$ to what power gives $8$? The answer is $3$, so $\log_2 8 = 3$. No calculator, two steps, and notice the pattern: combine the logs first with a law, then translate the tidy result back into a power.

Another. Simplify $2\log_a 3 + \log_a 5$. The power law runs in reverse here, pulling the $2$ back inside as an index: $2\log_a 3 = \log_a 3^2 = \log_a 9$. Then the product law combines the two logs: $\log_a 9 + \log_a 5 = \log_a 45$. Being able to run the power law in both directions, coefficient-to-index and index-to-coefficient, is the single most useful habit in the topic. The same fluency with indices that the surds and indices guide drills is precisely what makes these moves feel automatic.

Changing base

The TMUA can give you logs in a base that does not match the one you want, so you need to convert. The change of base formula is $\log_a x = \frac{\log_b x}{\log_b a}$, where $b$ is any base you find convenient. In words, a log in base $a$ becomes a ratio of two logs in your chosen base, with the old base sitting in the denominator.

This matters for two reasons. First, it lets you bring everything into a single base so the three laws apply across the whole expression. Second, a tidy special case falls out by swapping the roles: $\log_a b = \frac{1}{\log_b a}$. So $\log_2 8$ and $\log_8 2$ are reciprocals of each other, which is the kind of structural fact the examiners love to hide inside a longer question.

A worked example. Evaluate $\log_9 27$ without a calculator. Both $9$ and $27$ are powers of $3$, so change to base $3$: $\log_9 27 = \frac{\log_3 27}{\log_3 9} = \frac{3}{2}$, since $\log_3 27 = 3$ and $\log_3 9 = 2$. The trick is to spot a shared base lurking underneath and convert towards it, which turns an awkward log into a ratio of integers you can read off by sight. Hunting for that hidden common base is a recurring calculator-free move, and our guide to calculator-free techniques covers the same instinct across other topics.

Solving exponential equations

This is where logs earn their place. When the unknown sits in an exponent, as in $2^x = 7$, you cannot isolate $x$ by ordinary algebra because it is stuck upstairs. The standard method is to take logs of both sides and let the power law bring it down.

Taking logs of $2^x = 7$ gives $\log(2^x) = \log 7$, and the power law turns the left side into $x \log 2 = \log 7$. Now $x$ is a plain coefficient, so divide: $x = \frac{\log 7}{\log 2}$. That is the whole method. The base you take logs in does not matter, so most people use the natural log: $x = \frac{\ln 7}{\ln 2}$. The natural log $\ln x$ is just the log to base $e$, and it pairs with $e^x$ as its exact inverse, so $\ln(e^x) = x$ and $e^{\ln x} = x$. For equations built around $e$, taking $\ln$ of both sides is the cleanest possible route.

A slightly harder shape hides a quadratic. Solve $e^{2x} - 5e^x + 6 = 0$. The exponentials look intimidating until you substitute $y = e^x$, which turns the equation into $y^2 - 5y + 6 = 0$. That factorises as $(y - 2)(y - 3) = 0$, so $y = 2$ or $y = 3$. Undo the substitution: $e^x = 2$ or $e^x = 3$, giving $x = \ln 2$ or $x = \ln 3$. Spotting that an exponential equation is secretly a quadratic in disguise is a classic exam move, and it rewards the same simplify-first discipline the rest of the paper does.

Try one

Reading about it only takes you so far. Here is a real past-paper question that leans on exactly this machinery. Give it a genuine attempt against the clock before you reveal the solution, and watch for where a log law or an exponential substitution unlocks the whole thing:

If that felt slow, it is almost always because the log-as-index link has not fully clicked yet rather than any gap in the algebra, and that link is the easiest thing in the world to drill.

Common traps to avoid

A handful of errors account for most of the dropped marks here, and the TMUA writes answer options specifically to catch them.

• Splitting a log of a sum. $\log_a(x + y)$ is not $\log_a x + \log_a y$. The product law applies to a product inside the log, never to a sum. This is the single most common slip, and it is a favourite Paper 2 proof trap.
• Mishandling the power law. $(\log_a x)^2$ is not the same as $\log_a(x^2)$. The power law only moves an exponent that is on the number inside the log, not one on the whole log.
• Forgetting the domain. You can only take the log of a positive number, so $\log_a x$ requires $x > 0$. A negative or zero argument is a flag that you have gone wrong, and reasoning questions exploit this.
• Flipping change of base. Putting the new base on top instead of the bottom. Fix the convention: the old base goes in the denominator.
• Treating $\ln$ as a multiplier. $\ln(2x)$ is not $2\ln x$. It is $\ln 2 + \ln x$ by the product law. Reading $\ln$ as something you can factor out is a costly habit.

Most of these are structural errors, not arithmetic ones, which is good news: a few minutes spent on why each law holds inoculates you against all of them at once. The TMUA is hard mainly because of pace, and clean log work is one of the cheapest places to stop leaking time on avoidable slips.

How to drill it

You make this automatic the same way you make anything automatic: short, frequent, mixed practice rather than one marathon session.

Start by getting the link cold. Whenever you see $\log_a b$, force yourself to say "the power $a$ must be raised to, to get $b$" until the translation is instant in both directions. Then write the three laws from memory and test yourself on quick evaluations like $\log_2 32$, $\log_3 81$ and $\log_4 2$ with nothing written down. The target is not just the right answer, it is the right answer in seconds, because that is the speed the real paper demands.

Then practise the exponential-equation method until it is a reflex: logs of both sides, power law down, solve. Mix in the disguised-quadratic shape so you learn to spot when a substitution is hiding in plain sight. Because logs and exponentials lean so heavily on indices, every bit of index fluency you build feeds straight back in, which is why doing the two topics together pays off. A structured plan that sequences all of this is laid out in how to prepare for the TMUA. Get the log-as-index idea early and a whole class of questions stops being scary.

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