Solving Exponential Equations

You put £1000 in an account paying 5% a year. A natural question: how long until it doubles? After n years the balance is 1000 \times 1.05^{\,n}, so "doubled" means solving

1.05^{\,n} = 2.

Look where the unknown is: stuck up in the exponent. Ordinary algebra — adding, subtracting, multiplying, taking roots — can reach a base or a coefficient, but it can't get up there and pull n down. The same wall blocks every question of the same shape: when will the population reach a million? how old is this bone? how long until the medicine is half gone?

The key that unlocks all of them is the logarithm. A log is built to undo a power, and — through the laws of logarithms — it lets you slide the exponent down to ground level where you can reach it.

The method: take logs, drop the power, divide

Take the plainest version, 2^x = 20. We can't isolate x by rearranging — it isn't added or multiplied, it's an exponent. So take the logarithm of both whole sides, which keeps the equation balanced:

\log(2^x) = \log 20.

Now unleash the power law, \log(a^x) = x\log a, to bring the exponent down in front:

x\log 2 = \log 20.

With x back on the ground floor, it's an ordinary linear equation — just divide both sides by \log 2:

x = \frac{\log 20}{\log 2} \approx 4.32.

That's the whole recipe: take logs, drop the power, divide. It solves any b^x = k at all — the base of the log doesn't even matter, as long as you use the same one on both sides, because it cancels in the ratio.

See it solved

Step through the solution to 2^x = 20: take logs of both sides, use the power law to bring the exponent down, then divide.

Worked example 2 — read it off directly

Not every equation needs logs. Solve

2^x = 32.

You could grind through x = \log 32 / \log 2 — but notice 32 = 2^5. So the equation is really 2^x = 2^5: same base on both sides, so the exponents must match, and x = 5 by inspection. Whenever both sides are powers of the same base, rewrite them that way and just equate the exponents — no logarithm required.

The same trick handles hidden cases like 4^x = 8: write both as powers of 2, giving 2^{2x} = 2^3, so 2x = 3 and x = \tfrac{3}{2}.

Worked example 3 — the doubling-time question

Back to the money. £1000 growing at 5% a year doubles when 1.05^{\,n} = 2. Take logs of both sides and drop the power:

n\log 1.05 = \log 2 \quad\Rightarrow\quad n = \frac{\log 2}{\log 1.05} \approx 14.2 \text{ years}.

So it takes a little over 14 years to double — and, beautifully, the answer doesn't depend on how much you started with. Change the rate to 7% and you'd solve 1.07^{\,n} = 2 instead, giving about 10.2 years. Every "how long until it doubles / triples / reaches a target" problem is exactly this: an unknown trapped in an exponent, freed by a logarithm.

Two classic slips sink most attempts at these:

Take the log of both WHOLE sides — then apply the power law, then divide. To bring an exponent down you must log the entire side and use \log(a^x) = x\log a. Logging only part of a side, or writing x = \log 20 and forgetting to divide by \log 2 at the end, is the number-one error. There is no "\log of a term" you can sprinkle on selectively — the operation acts on the whole side or the balance breaks.
Same base? Don't reach for logs at all. Faced with 2^{x+1} = 2^{3x-1}, many students dutifully take logs — but the bases already match, so you can equate the exponents directly: x+1 = 3x-1, giving x = 1. Reaching for logs still works, but it's slower and invites arithmetic mistakes. Spot the matching base first.

Worked example 4 — reaching a target population

A town of 40,000 people grows by 3% a year. When will it first pass 100,000? The population after n years is 40000 \times 1.03^{\,n}, so we need

40000 \times 1.03^{\,n} = 100000.

First get the exponential term alone — divide both sides by 40,000: 1.03^{\,n} = 2.5. Now take logs and drop the power:

n = \frac{\log 2.5}{\log 1.03} \approx 31 \text{ years}.

Notice the shape: isolate the power first, then take logs. If a coefficient is multiplying the exponential, clear it before you log — otherwise the power law has nothing clean to act on.

Not a bit — and that surprises people. Your calculator has both \log (base 10) and \ln (base e), and either solves 2^x = 20 to the same answer, because x = \dfrac{\log 20}{\log 2} = \dfrac{\ln 20}{\ln 2} \approx 4.32. The base cancels in the ratio. So use whichever you like — just be consistent and use the same base top and bottom. Scientists usually reach for \ln because it dovetails with calculus and with continuous growth e^{kt}; exam mark schemes are happy with either.

Yes — and that's the quiet power of this one idea. A radioactive sample decays as N = N_0\,(\tfrac12)^{t/h}; to find its half-life or its age you must unstick t from the exponent — a logarithm. That's literally how archaeologists carbon-date an ancient bone: measure how much carbon-14 is left, then solve an exponential equation for the time elapsed. Bacteria, viral spread and compound interest run the same film in reverse (doubling times instead of half-lives), and the maths is identical.

There's even a famous shortcut: the rule of 72. Divide 72 by the percentage growth rate to estimate the doubling time — 72 ÷ 5 ≈ 14.4 years, spot on our earlier answer. Why 72? Because the exact answer is \ln 2 / \ln(1+r) \approx 0.693/r, and 69.3 rounded up to the friendlier, more-divisible 72. The rule of 72 is a logarithm wearing a disguise.

See it explained

Sal Khan solves an exponential equation by taking logarithms of both sides.