- An exponential equation has the unknown in the power, as in \( 1.06^n = 3 \). Taking logarithms of both sides brings the unknown down: \( n\ln 1.06 = \ln 3 \).
- For \( a \cdot b^t = c \), the solution is \( t = \dfrac{\ln(c/a)}{\ln b} \). For continuous growth \( a\,e^{kt} = c \), it is \( t = \dfrac{\ln(c/a)}{k} \).
- Case: an online shop whose orders grow 6% a month takes about 18.85 months to triple, so it first reaches triple its current orders in month 19.
- The Rule of 70 estimates doubling time as \( 70 \div \) (growth rate in %). At 6% it gives 11.67 months against an exact 11.90.
- Exam point: keep logarithms exact until the final line, and round time answers in the direction the context needs.
The case: when will the orders triple?
A small online shop selling handmade candles has had a good year. It now handles 2,400 orders a month, and for the past year the number has grown by roughly 6% a month. The owner rents a small unit, and her landlord has told her it can cope with at most 7,200 orders a month, three times today’s level. Beyond that she will need a bigger warehouse, and leases in her area need about six months’ notice. The numbers are illustrative, not data from a real business.
If growth carries on at 6% a month, orders after \( n \) months will be \( 2400 \times 1.06^n \). Her question: in how many months will orders reach 7,200, and when does she need to start looking for a new warehouse? We need to solve
Trial and error would get there slowly: \( 1.06^{10} \approx 1.79 \), \( 1.06^{20} \approx 3.21 \), so the answer is somewhere between 10 and 20 months. To get it exactly, we need a way to undo the power.
The concept: logarithms undo powers
Addition is undone by subtraction, multiplication by division, and squaring by square roots. Raising a fixed number to a power is undone by a logarithm. The statement \( \log_b y = x \) means exactly the same as \( b^x = y \): the logarithm answers the question “what power of \( b \) gives \( y \)?” So \( \log_{10} 1000 = 3 \) because \( 10^3 = 1000 \).
The property that makes logarithms so useful for equations is the power law, \( \ln(b^n) = n\ln b \). Taking the log of a power turns the exponent into an ordinary multiplier. An unknown stuck up in the power comes down to the main line, where normal algebra can deal with it.[1]
Any base of logarithm works, as long as you use the same base on both sides. In practice we use the natural logarithm, \( \ln \), which is \( \log_e \), because it is on every calculator and it fits continuous growth, \( e^{kt} \), perfectly: \( \ln(e^{kt}) = kt \).
The formula
Here \( a \) is the starting value, \( c \) is the target value, \( b \) is the growth factor per period (1.06 for 6% growth, 0.9 for 10% decay), \( k \) is the continuous growth rate (negative for decay), and \( t \) is the number of periods. Both formulas need \( a \) and \( c \) to have the same sign, and \( b > 0 \), \( b \neq 1 \), \( k \neq 0 \).
Two special cases come up constantly. The doubling time is the case \( c/a = 2 \), so \( T_2 = \dfrac{\ln 2}{\ln b} \). The half-life is the case \( c/a = \tfrac{1}{2} \), so \( T_{1/2} = \dfrac{\ln 2}{|k|} \) for continuous decay. Neither depends on the starting amount.
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Why this method, and when it is not needed
The method works because \( \ln \) is a one-to-one function: if two positive numbers are equal, their logarithms are equal, and the other way round. So taking \( \ln \) of both sides never creates or loses a solution, provided both sides are positive. Then the power law does the real work.
It fits our case because the shop’s growth is multiplicative: every month, orders are multiplied by the same factor, 1.06. Equal steps in time give equal ratios, not equal differences. That is the signature of exponential change, and logarithms are the natural tool for undoing it. If orders instead grew by a fixed 150 a month, the model would be linear, \( 2400 + 150n \), and simple algebra would do.
You also do not need logs when both sides can be written as powers of the same base. For \( 4^x = 32 \), write \( 2^{2x} = 2^5 \), so \( 2x = 5 \) and \( x = 2.5 \). Logs give the same answer, but matching the bases is quicker and keeps the answer exact.
Setting it up
Before taking any logs, get the power on its own:
- Starting value \( a = 2400 \) orders a month.
- Growth factor \( b = 1 + 0.06 = 1.06 \) per month. (Using \( b = 0.06 \) or \( b = 6 \) is a classic slip.)
- Target \( c = 7200 \), so \( c/a = 3 \).
- Unknown \( n \), in months, with \( n \ge 0 \). Orders are counted month by month, so in the end we will need a whole number.
Dividing first, to get \( 1.06^n = 3 \), avoids the tempting but wrong step of taking \( \ln \) of \( 2400 \times 1.06^n \) as if it were \( 2400 \times \ln 1.06^n \).
Solving it
Solve \( 2400 \times 1.06^n = 7200 \), and find the first whole month in which orders reach 7,200.
With six months’ notice needed, the owner should start looking for a new unit within about 12 to 13 months from now, and sooner if growth speeds up.
The same technique handles decay, where the power has a negative rate. The next example uses continuous decay in base \( e \), which is how drug and caffeine levels in the body are usually modelled.
A strong coffee at 8 am puts about 200 mg of caffeine into the bloodstream. Assume the amount left decays continuously with a half-life of 5 hours. (Real half-lives vary a lot from person to person; 5 hours is a convenient assumption.) At what time will the amount fall to 30 mg?
So the level falls to 30 mg at about 9:41 pm. Both logs in Step 4 are negative (\( \ln 0.15 < 0 \) and \( k < 0 \)), so the time comes out positive, as it should.
Exam questions often put the unknown in the power on both sides, with different bases. Logs still work: take them of both sides, expand with the power law, and collect the terms in \( x \).
Exam style. Solve \( 5^{x+1} = 2^{3x} \), giving \( x \) to 3 significant figures.
Check: \( 5^{4.424} \approx 1240 \) and \( 2^{10.27} \approx 1240 \). The two sides agree.
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What the answer tells us
An answer of 18.85 months means that orders pass 7,200 part-way through month 19 of growth. Because orders are counted monthly, the useful answer is a whole number, and because the question is “when do we hit the limit?”, we round up. Rounding down to 18 would tell the owner she is safe in a month when she is not.
The answer is reasonable: our first guesses showed it lies between 10 and 20 months. It also lines up with a well-known shortcut. Doubling takes \( \dfrac{\ln 2}{\ln 1.06} \approx 11.90 \) months, and tripling takes \( \dfrac{\ln 3}{\ln 2} \approx 1.585 \) times as long as doubling, and indeed \( \dfrac{\ln 2}{\ln 1.06} \times \dfrac{\ln 3}{\ln 2} = \dfrac{\ln 3}{\ln 1.06} \approx 18.85 \).
The Rule of 70 says doubling time is roughly 70 divided by the percentage growth rate. It comes from \( \ln 2 \approx 0.693 \) and \( \ln(1 + g) \approx g \) for small \( g \).[2] The table shows how good the rule is, and how sensitive the tripling time is to the growth rate.
| Monthly growth | Exact doubling time (months) | Rule of 70 | Exact tripling time (months) |
|---|---|---|---|
| 3% | 23.45 | 23.33 | 37.17 |
| 6% | 11.90 | 11.67 | 18.85 |
| 9% | 8.04 | 7.78 | 12.75 |
| 12% | 6.12 | 5.83 | 9.69 |
The rule is very close at low rates and slightly underestimates at higher ones, because \( \ln(1 + g) \) is a little less than \( g \). The bigger lesson is sensitivity: if growth slips from 6% to 3% a month, the owner has twice as long; if it climbs to 12%, she has under 10 months. If the 6% were instead a continuous rate, \( 2400e^{0.06t} = 7200 \) would give \( t = \dfrac{\ln 3}{0.06} \approx 18.31 \) months, a little sooner, because continuous growth compounds faster. Figure 1 shows the monthly model and the warehouse limit.
For the full story of exponentials and logarithms, including graphs, laws of logs and linearising data, see Chapter 5 of the Mathematics Made Simple textbook. To build speed, the Mathematics Made Simple practice book has exponential and log equations at every level, each with full working.
Get both books in the Mathematics Made Simple bundle for $16 and save $5 on your exponentials revision.Both books · $16
Practice questions
EasyQ1. Solve \( 2^x = 20 \), giving \( x \) to 3 decimal places. Show worked answer
\( x = \dfrac{\ln 20}{\ln 2} = \dfrac{2.995732}{0.693147} \approx 4.322 \).
EasyQ2. Solve \( e^{3x} = 12 \), giving \( x \) to 3 decimal places. Show worked answer
\( x = \dfrac{\ln 12}{3} = \dfrac{2.484907}{3} \approx 0.828 \).
MediumQ3. $5,000 is invested at 4% interest, compounded annually. After how many complete years will it first be worth more than $8,000? Show worked answer
\( n > \dfrac{\ln 1.6}{\ln 1.04} \approx 11.98 \).
So 12 complete years. Check: \( 5000 \times 1.04^{11} \approx \$7697.27 \) and \( 5000 \times 1.04^{12} \approx \$8005.16 \).
MediumQ4. Solve \( 3^{2x}-4 \times 3^x + 3 = 0 \). Show worked answer
\( (y-1)(y-3) = 0 \), so \( y = 1 \) or \( y = 3 \).
\( 3^x = 1 \) gives \( x = 0 \); \( 3^x = 3 \) gives \( x = 1 \).
ExamQ5. The mass of a radioactive sample is \( m = m_0e^{-kt} \), with \( t \) in hours. After 3 hours, 40% of the original mass has decayed. (a) Show that \( k = \tfrac{1}{3}\ln\tfrac{5}{3} \) and give \( k \) to 4 decimal places. [2] (b) Find the half-life, in hours, to 3 significant figures. [2] (c) Find how long it takes until less than 5% of the original mass remains, to 3 significant figures. [3] Show worked answer
(b) \( e^{-kT} = \tfrac{1}{2} \) gives \( T = \dfrac{\ln 2}{k} \) [1] \( = \dfrac{3\ln 2}{\ln(5/3)} \approx 4.07 \) hours [1].
(c) \( e^{-kt} < 0.05 \) [1], so \( -kt < \ln 0.05 \) and \( t > \dfrac{\ln 20}{k} \) [1].
\( t > \dfrac{3\ln 20}{\ln(5/3)} \approx 17.6 \) hours [1].
Frequently asked questions
Should I use ln or log base 10 to solve exponential equations?
Either, as long as you use the same one on both sides: \( \dfrac{\log 3}{\log 1.06} \) and \( \dfrac{\ln 3}{\ln 1.06} \) give the same 18.85. Natural logs are the better habit, because they also handle \( e^{kt} \) in one step.
What is the Rule of 70 and how accurate is it?
The Rule of 70 estimates doubling time as 70 divided by the percentage growth rate per period. It is accurate to within a few per cent for growth rates up to about 10%, and gets steadily worse for higher rates. Some people use 72 instead, because 72 has more factors and it is a little more accurate around 8%.
Why can’t I take the log of a negative number?
Because no real power of a positive base gives a negative result, so \( \ln x \) is only defined for \( x > 0 \). An equation like \( 2^x = -4 \) therefore has no real solution, and in a disguised quadratic you reject any root with \( 3^x < 0 \).
Does the doubling time depend on the starting amount?
No. In \( a \cdot b^T = 2a \), the \( a \) cancels and leaves \( b^T = 2 \). A shop with 240 orders and a shop with 24,000 orders, both growing 6% a month, double in the same 11.90 months.
References
- Abramson, J. (2017). Precalculus. Houston, TX: OpenStax. Chapter 4, Exponential and Logarithmic Functions. https://openstax.org/details/books/precalculus
- Chiang, A. C. and Wainwright, K. (2005). Fundamental Methods of Mathematical Economics, 4th edition. New York: McGraw-Hill.
- Cambridge International. Cambridge International AS & A Level Mathematics 9709: Syllabus for examination in 2026 and 2027. Cambridge: Cambridge University Press & Assessment.
- International Baccalaureate Organization (2019). Mathematics: analysis and approaches guide (first assessment 2021). Cardiff: IBO.
- College Board (2023). AP Precalculus Course and Exam Description. New York: College Board.
