You cannot tell from looking which method it needs

the hidden quadratic

2real roots

substitutionwhat solves it

exactanswer

Three equations that look nothing like their difficulty. The method comes from the structure, not the appearance.

The figure needs JavaScript. It shows three equations: e^2x − 5e^x + 4 = 0, whose substitution u = e^x gives u = 1 and 4 and so x = 0 and x = 1.386; x⁴ + 5x − 6 = 0, which factorises to roots 1 and −2 with the remaining quadratic having discriminant −11; and e^x = sin x, which has no positive solutions and infinitely many negative ones, the largest near −3.183. Everything below stands on its own.

The hidden quadratic

e^2x − 5e^x + 4 = 0 is not an exponential equation to be attacked with logarithms. It is a quadratic wearing a disguise, because e^2x = (e^x)². Put u = e^x:

u² − 5u + 4 = 0, so (u − 1)(u − 4) = 0 and u = 1 or 4

Then undo the substitution, which is the step people forget:

e^x = 1 gives x = 0.
e^x = 4 gives x = ln 4 = 1.386, or exactly 2 ln 2.

Two roots, and the first is the one that gets dropped, because ln 1 = 0 looks like nothing happening. Both are real answers and both are marks.

The signal to look for is a squared version of something alongside the thing itself: e^2x with e^x, or 4^x with 2^x, or x⁴ with x². Whenever you see it, substitute.

Not every u comes back. If a substitution gives u = −3 then e^x = −3 has no solution, because an exponential is never negative, and that u is discarded. Say so explicitly: "e^x = −3 has no real solutions, so this root is rejected" is the sentence markschemes want, and it is a mark on its own.

The one that looks worse than it is

x⁴ + 5x − 6 = 0 has a fourth power and no obvious factor pairs. Try the small integers anyway:

at x = 1: 1 + 5 − 6 = 0, and at x = −2: 16 − 10 − 6 = 0

so (x − 1) and (x + 2) are both factors, and dividing out leaves

x⁴ + 5x − 6 = (x − 1)(x + 2)(x² − x + 3)

The last bracket has discriminant 1 − 12 = −11, so it contributes no real roots. Exactly two real roots, both integers, from an equation that looked like it needed a machine.

Always try x = 1, −1, 2 and −2 first. It takes fifteen seconds and it is the difference between an exact answer and a decimal.

The one that looks better than it is

e^x = sin x has two functions you know well and no route at all. There is no rearrangement that isolates x, and that is not a gap in your technique; no such rearrangement exists.

What you CAN say without a calculator:

No solutions for x ≥ 0. There, e^x ≥ 1 while sin x ≤ 1, and they are equal only if both are exactly 1, which happens at different places. At x = 0 the two sides are 1 and 0.
Infinitely many for x < 0, because e^x shrinks towards zero while sin x keeps oscillating between −1 and 1, so it crosses the shrinking curve twice every cycle, for ever: once as the sine rises past it and once as the sine falls back.
The largest of them is near x = −3.183, where both sides are about 0.041.

So the honest answer to "solve e^x = sin x" is a graph, a root from the machine, and a sentence saying there are infinitely many more. Stating how many there are is part of the answer.

How to choose, in thirty seconds.

Is something squared alongside itself? Substitute.
Is it a polynomial? Try the small integers, then factorise.
Can you get one base on both sides? Equate the exponents.
Is the unknown in an exponent on its own? Take logs.
None of those? Graph it, and say how many solutions there are.

Run the list in order and the last line is a decision rather than a surrender.

On the GDC: when the machine is the method

This is the one sub-topic where the calculator is part of the syllabus content rather than a convenience: the guide explicitly includes equations with no appropriate analytic approach. The skill is reading how many solutions there are, not just finding one.

When you may use it. Analysis Paper 1 is non-calculator, so the equations there are the ones with structure: the hidden quadratic and the polynomial with integer roots. Paper 2 and Paper 3 are where e^x = sin x belongs.

TI-Nspire CX II

Graph both sides separately rather than the difference: f1(x)=e^x and f2(x)=sin(x), in radians: doc → Settings → Document Settings → Angle → Radian. In degrees the sine flattens to almost nothing and the figure loses every crossing Use the x² key or the right arrow to leave the exponent: typing ^ opens a superscript box and everything after it stays inside.
menu → Window / Zoom → Window Settings, x from -12 to 2, y from -1.5 to 2
Analyze Graph → Intersection, bounded either side of each crossing: -3.183, then -6.28, and they keep coming
nSolve(e^x=sin(x),x=-4) also works, and the x=-4 is a starting guess, which is why it lands on the crossing near −3.183. Written as a bare third argument, ,-4), it is read as a LOWER BOUND instead, so asking for −9 that way still hands back −3.183

Casio fx-CG50

SHIFT MENU SET UP and set Angle to Rad first, or the sine curve is the wrong shape entirely
MENU → Graph, with Y1=e^X and Y2=sin X, then F6 DRAW
SHIFT F5 G-SOLVE → ISCT, which reports the leftmost crossing in the window first; the right arrow then walks right through them one at a time
MENU → Equation → Solver finds one root near a guess, which is less useful here than seeing the pattern

The mark people lose. Reporting one root when the question asked you to solve the equation. A solver returns a single answer and says nothing about how many there are. Graph it, count the crossings in the interval you were given, and if there are infinitely many, say so. On e^2x − 5e^x + 4 = 0 the same habit catches the dropped x = 0: two values of u mean two equations to solve, not one.

Where this actually turns up

The content list ends with applications to real situations, and it links back to exponential growth at 2.9. The useful thing is that the two halves of this page are both there: some contexts solve exactly, and some do not, and the context never tells you which.

Exactly. A 200 mg dose of caffeine has a half-life of about 5 hours, so the amount left after t hours is 200 × 0.5^t/5. When is it down to 50 mg?

200 × 0.5^t/5 = 50, so 0.5^t/5 = 0.25, and 0.25 is 0.5², so t/5 = 2 and t = 10 hours. No calculator needed and the answer is a whole number, because the question was built from two half-lives.

Not exactly. A car worth £15,000 loses 15% of its value a year, so it is worth 15000 × 0.85^t. Repairs run at about £400 a year, so the total spent is 400t. When has the owner spent as much on repairs as the car is now worth?

15000 × 0.85^t = 400t. There is no way to rearrange this. The t is in an exponent on one side and on its own on the other, and no logarithm separates them. Graph both and read the crossing: t = 8.87 years, where each side is £3548.

Both of those are "exponential" questions and they need different tools. The first has the unknown in one place, so a logarithm reaches it. The second has it in two, so nothing does. That is the same test as the rest of this page, applied to a context rather than to a bare equation: count how many places the unknown appears before you choose a method.

Your turn

1. Solve e^2x − 5e^x + 4 = 0. Give the smaller root.

2. And the larger root, to 3 decimal places.

3. How many real roots does x⁴ + 5x − 6 = 0 have?

4. How many solutions does e^x = sin x have with x ≥ 0?

5. What tells you to substitute u = e^x?

Question 5. What tells you to substitute u equals e to the x? The equation has an e in it. e^2x is (e^x)², so the equation is a quadratic in e^x. There is nothing else to try. The constant term is 4.

Where the marks go

1 markThe substitution or the factorisation, stated.

1 markThe values of the new variable.

1 markUndoing the substitution, both ways.

1 markRejecting any impossible value, with a reason.

The third and fourth marks are where this sub-topic is won and lost. A student who finds u = 1 and u = 4 and reports only ln 4 has done the difficult part and lost two of the four.