May 20, 20269 min read

Exponential & Logarithmic Functions in Pre-Calculus

Logarithms have a reputation for being arbitrary — a button on the calculator, a set of laws to memorise, a unit to survive. Students get through the chapter by pattern-matching and quietly hope the topic does not come back.

It comes back. It is in Calculus 12, in first-year calculus, in chemistry, and in every compound-interest question anyone will ever ask you. And the reputation is undeserved, because underneath there is exactly one idea.

That idea takes about a minute. If this unit is going badly, it is almost always because that minute got skipped — which is what Pre-Calculus 12 tutoring tends to start with.

The one idea: a logarithm is an exponent

That is the whole thing. A logarithm is not a new kind of operation. It is the answer to a question you already know how to ask.

2^{3} = 8

Now ask the same fact backwards. "What power do I raise 2 to, in order to get 8?" The answer is 3, and the notation for that question is:

lo g_{2} 8 = 3

Identical fact. Same three numbers, same relationship, rearranged. So any log statement can be rewritten as an exponent statement and vice versa:

lo g_{b} x = y ⟺ b^{y} = x

If you take one thing from this page, take that line. Written on the top of the page in an exam, it turns most of the unit into translation. Students who are stuck on a log question are almost always stuck because they are trying to think about logs; the ones who are fine have quietly converted it to an exponent and are looking at something familiar.

Two consequences drop straight out and they are worth knowing on sight:

b^{l o g_{b} x} = x and lo g_{b} (b^{x}) = x

They look like rules to learn. They are just the sentence "these two operations undo each other", written in symbols. Raise 2 to the power that gets you 8, and you get 8. Of course you do.

Why anyone invented them

Logarithms exist to solve for an exponent. That is the job, and nothing else does it.

3^{x} = 20

Try to solve that with the algebra you already have and you cannot. You cannot divide by 3 — x is not a factor, it is an exponent, and division does not reach it. Every tool from earlier years fails, because the unknown is in a place none of them can touch.

So take the log of both sides, use the power law, and the exponent comes down where you can get at it:

x ln 3 = ln 20 ⟹ x = \frac{ln 20}{ln 3} = 2.727

Check it: $3^{2.727} = 20$ . ✓

That move — bring the exponent down — is what logarithms are for. Everything else in the unit is support for that one manoeuvre.

The picture: two curves, one relationship

"The logarithm is the inverse of the exponential" is a sentence students repeat and mostly do not believe. Here is the version you can see.

Two curves, one relationship. Every point on one has a twin on the other with the coordinates swapped — (2, 4) becomes (4, 2). That is what "inverse" means, and it is why log₂x is undefined for x ≤ 0: the blue curve never gets there, so the orange one has nothing to reflect.

The two curves are mirror images across the dashed line. Fold the page along y = x and they land on each other exactly. Every point has a twin with its coordinates swapped: (0, 1) becomes (1, 0); (2, 4) becomes (4, 2).

And this explains the domain

Students are told $lo g_{2} x$ is undefined for $x \leq 0$ and dutifully accept it. The picture explains why, in one glance.

Look at the blue curve. $2^{x}$ is never zero and never negative — it gets small, it flattens toward the axis, it never touches it. Ask for $2^{x} = - 8$ and there is no answer, because no power of 2 is negative. So when you reflect that curve, the orange one has nothing to reflect at $x \leq 0$ . The hole in the log's domain is the exponential's range, mirrored.

This is not a rule that had to be memorised. It is a fact about the picture, and it is why the domain check at the end of a log question is not bureaucracy — it is the only thing standing between you and an answer that does not exist.

The laws, and the one that is not a law

Three laws, and each is an exponent rule wearing a disguise.

lo g (ab) = lo g a + lo g b

lo g (\frac{a}{b}) = lo g a - lo g b

lo g (a^{n}) = n lo g a

Check the first: log(4 × 25) = log 100 = 2, and log 4 + log 25 = 0.602 + 1.398 = 2. ✓ And notice what it is really saying — multiplying the numbers adds their logs. That is the entire reason logarithms were invented in the seventeenth century: they turn multiplication into addition, which before calculators was the difference between an afternoon and a minute.

Now the one that ends more Pre-Calculus 12 questions than any other:

lo g (a + b) \neq = lo g a + lo g b

Test it. $lo g 20 = 1.301$ , while $lo g 10 + lo g 10 = 2$ . Not close. The real law is about products; there is no law at all for sums, and log(a+b) simply does not break apart. If you find yourself splitting a log across a plus sign, stop — you are not doing algebra, you are inventing it.

This belongs to a bigger family of errors worth knowing by name. We catalogued them in the ten mistakes that cost the most marks in Pre-Calculus 12.

What is e, anyway?

We have been writing $ln$ without saying what it is. $ln x$ is just $lo g_{e} x$ — a logarithm like any other, with an odd-looking base of 2.71828…

Which raises the fair question students rarely get answered: where does that number come from, and why would anyone choose it? It is not arbitrary, and the answer is sitting in the compound interest we were just doing.

Put $1 in an account paying 100% a year. Compounded once, you end with $2. Now compound it more often — the rate is split, but it is applied more times:

Quarterly: (1 + 1/4)⁴ = $2.4414
Monthly: (1 + 1/12)¹² = $2.6130
Daily: (1 + 1/365)³⁶⁵ = $2.7146
Hourly: (1 + 1/8760)⁸⁷⁶⁰ = $2.7181

Compounding faster earns more — but look at the numbers. They are slowing down. They are closing on something rather than running away, and no matter how often you compound, you never reach $2.72:

n \to \infty lim (1 + \frac{1}{n})^{n} = e = 2.71828 \dots

That is e. It is what continuous growth converges to — the answer to "what if the compounding never stops?" And the striking part is that it is finite. Infinitely frequent compounding does not give infinite money; it gives $2.72. Nature has a speed limit on that kind of growth, and e is the number at it.

Which is why e is the natural base rather than a mathematician's affectation. Anything growing in proportion to its own size — money, bacteria, a decaying isotope, a cooling cup of coffee — is doing this, and e is the constant that falls out of it.

Change of base

Your calculator offers log (base 10) and ln (base e), and the question wants base 2. So convert:

lo g_{b} x = \frac{ln x}{ln b}

So $lo g_{2} 10 = ln 10/ ln 2 = 3.322$ , and checking, $2^{3.322} = 10$ . ✓ The base on the bottom is the base you want. Get that upside down and every answer is a plausible-looking reciprocal.

Where this is actually used

Two examples, because this is one of the rare Pre-Calculus topics where the applications are not contrived.

How long to double your money

At 5% a year, $1.0 5^{t} = 2$ . Solve for the exponent:

t = \frac{ln 2}{ln 1.05} = 14.2 years

Notice what is not in that calculation: the amount of money. It cancels. $1,000 and $100,000 both take 14.2 years to double at 5%, which is the sort of fact that sounds wrong and is not. (Bankers use "70 divided by the rate" as a shortcut — 70/5 = 14, and now you know where their 70 comes from.)

Carbon dating

Carbon-14 has a half-life of 5,730 years, so the fraction left after t years is $(1/2)^{t /5730}$ . A sample with 25% remaining:

t = 5730 \cdot \frac{ln 0.25}{ln 0.5} = 11, 460 years

Which is two half-lives, as it must be — a good sign the algebra is right. But the method handles the awkward numbers too: 70% remaining gives about 2,950 years, and no amount of halving-in-your-head gets you that. That is the point of having the equation.

Two to try

1. The domain trap

lo g_{2} x + lo g_{2} (x - 2) = 3

Combine using the product law, then convert to exponent form:

lo g_{2} (x (x - 2)) = 3 ⟹ x^{2} - 2 x = 2^{3} = 8

x^{2} - 2 x - 8 = 0 ⟹ (x - 4) (x + 2) = 0

So x = 4 or x = −2. Check them in the $original$ . x = 4: $lo g_{2} 4 + lo g_{2} 2 = 2 + 1 = 3$ ✓. x = −2: that needs $lo g_{2} (- 2)$ , which does not exist. ✗

Answer: x = 4 only. This question exists to check whether you discard the second root, and roughly half of students do not. The algebra was never the test.

2. Solve for the exponent

A population of 500 grows 8% a year. When does it pass 1,200? Set $500 (1.08)^{t} = 1200$ , so $1.0 8^{t} = 2.4$ , and $t = ln 2.4/ ln 1.08 = 11.4$ years. Same move every time: isolate the power, then log both sides.

Where marks actually go

Splitting a log across a plus sign. Costs the question.
Keeping a root that the domain forbids.
Change of base upside down.
Forgetting that the base must be positive and not 1 — a fact that looks pedantic until a question tests it.
Trying to solve for an exponent with ordinary algebra, getting stuck, and assuming the question is broken. It is not; you just have not reached for the one tool that works.

Where it goes next

Exponentials do not stop at Pre-Calculus 12 — they get more important. In calculus, the derivative of eˣ is eˣ, the only function that is its own rate of change, and that single property is why e turns up in every growth and decay model there is. Students who leave Grade 12 treating logs as a button struggle there, and it looks like a calculus problem.

Getting help with Pre-Calculus 12

If this unit is not landing, the fix is usually smaller than it feels — most students are missing the one translation line at the top of this page, and everything downstream is a consequence. Our one-on-one Pre-Calculus 12 help starts by finding which line is missing rather than re-teaching the chapter.

Sessions run in person in Burnaby or online across Metro Vancouver, aligned to the BC curriculum. Book a free 30-minute consultation and bring a recent test.