Solving exponential growth equations comes down to one core skill: using logarithms to pull a variable out of an exponent. Whether you’re finding how long an investment takes to double or calculating a bacterial growth rate, the algebra follows the same pattern. Isolate the exponential term, take the logarithm of both sides, then solve the resulting linear equation.
The Two Standard Formulas
Most exponential growth problems use one of two formulas. The first is the discrete growth model: y = a(1 + r)t, where a is the starting amount (the value when t = 0), r is the growth rate expressed as a decimal, and t is time. If a population of 500 grows at 3% per year, you’d write y = 500(1.03)t. The quantity (1 + r) is called the base. When the base is greater than 1, you have growth. When it’s between 0 and 1, you have decay.
The second is the continuous growth model: A = Pert, where P is the starting amount, r is the growth rate, t is time, and e is Euler’s number (approximately 2.718). This version is used whenever growth happens smoothly rather than in discrete steps, which makes it common in biology, physics, and continuously compounded finance. The two formulas are mathematically related, so the solving techniques are nearly identical.
Solving for Time
This is the most common type of problem. You know the starting amount, the growth rate, and the target amount, and you need to find how long it takes to get there. Here’s the process using a concrete example.
Problem: You invest $2,000 at 5% annual growth. How many years until it reaches $5,000?
Start with the equation: 5000 = 2000(1.05)t. Your first move is to isolate the exponential term by dividing both sides by 2000:
2.5 = (1.05)t
Now the variable is trapped in the exponent. To free it, take the logarithm of both sides. You can use either the natural log (ln) or the common log (log); both work. Using natural log:
ln(2.5) = ln(1.05t)
The logarithm power rule lets you pull the exponent down in front: ln(2.5) = t · ln(1.05). Now it’s a simple linear equation. Divide both sides by ln(1.05):
t = ln(2.5) / ln(1.05) = 0.9163 / 0.04879 ≈ 18.78 years
That’s the entire method. Every “solve for time” problem follows these same four steps: isolate the exponential, take the log of both sides, use the power rule to bring down the exponent, then solve the resulting equation with basic algebra.
When the Formula Uses e
If the problem uses continuous growth, the steps are even simpler because the natural log cancels e directly. Say a bacteria colony starts at 1,000 cells and grows continuously at a rate of 0.35 per hour. How long until there are 50,000 cells?
50000 = 1000 · e0.35t
Divide both sides by 1000: 50 = e0.35t
Take the natural log of both sides: ln(50) = 0.35t
Notice that ln and e cancel each other immediately, so you skip the power rule step entirely. Solve for t:
t = ln(50) / 0.35 = 3.912 / 0.35 ≈ 11.18 hours
Solving for the Growth Rate
Sometimes you know the starting value, the ending value, and how much time passed, but you need to find the rate. The algebra is the same, just rearranged.
Problem: A city’s population grew from 120,000 to 185,000 over 10 years. What was the annual growth rate?
185000 = 120000(1 + r)10
Divide both sides by 120000: 1.5417 = (1 + r)10
To undo the exponent, raise both sides to the power of 1/10 (the reciprocal of the exponent):
(1.5417)1/10 = 1 + r
1.0443 = 1 + r
r = 0.0443, or about 4.43% per year
You can also solve this with logarithms (take the log of both sides, divide by 10, then raise e or 10 to that power), but the reciprocal exponent method is faster and avoids an extra step. This formula works for any compound growth rate calculation: divide the ending value by the starting value, then raise the result to 1 divided by the number of periods.
Solving for the Starting Amount
This is the easiest version. Since the starting amount a isn’t in the exponent, no logarithms are needed at all. Just divide both sides by the growth factor.
Problem: After 6 years of 7% annual growth, an account holds $12,000. What was the original deposit?
12000 = a(1.07)6
Calculate (1.07)6 = 1.5007. Then divide: a = 12000 / 1.5007 ≈ $7,996.
Handling Compound Interest Frequencies
Financial problems often compound more than once per year, which adds one more variable. The formula becomes A = P(1 + r/n)nt, where n is the number of times interest compounds per year: 1 for annually, 4 for quarterly, 12 for monthly, 365 for daily. The exponent becomes nt (the number of compoundings per year multiplied by the number of years), and the rate inside the parentheses becomes r/n.
For example, $1,250 invested at 3% compounded monthly simplifies to A = 1250(1.0025)12t. The base is 1.0025 (that’s 0.03 divided by 12, then added to 1), and the exponent is 12t. If you need to solve for t, the logarithm method works exactly the same way. You’d isolate the exponential term, take the log of both sides, bring down the 12t using the power rule, and divide.
As the number of compounding periods increases, the result approaches the continuous formula A = Pert. This is why continuous compounding is common in theoretical problems: it’s actually simpler to work with, since you only need the natural log and never have to track compounding frequency.
The Rule of 72 for Quick Estimates
Before reaching for a calculator, you can estimate doubling time with a simple shortcut: divide 72 by the percentage growth rate. At 6% growth, an amount doubles in roughly 72 / 6 = 12 years. At 9% growth, it doubles in about 8 years.
This approximation is most accurate for growth rates between 6% and 10%, where it’s nearly exact at 8%. For lower rates around 2%, dividing 70 by the rate gives a better answer. For continuous compounding, 69.3 is the most precise numerator (because the natural log of 2 is 0.693), but 70 is close enough for mental math. These shortcuts are useful for checking whether your full calculation landed in the right ballpark.
A Real-World Example: Bacterial Growth
Exponential growth equations aren’t just textbook exercises. Microbiologists use the continuous model to track how fast bacteria multiply. E. coli growing on a rich glucose medium has a specific growth rate of about 1.26 per hour, which translates to a doubling time of roughly 33 minutes. On a poor carbon source like acetate, the growth rate drops to 0.21 per hour, and the doubling time stretches to about 3 hours and 18 minutes.
The doubling time formula comes directly from solving the continuous growth equation. Starting from X(t) = X₀ekt, set X(t) equal to 2X₀ (double the starting amount), then solve for t:
2 = ekt
ln(2) = kt
t = ln(2) / k = 0.693 / k
Plug in k = 1.26, and you get t = 0.693 / 1.26 = 0.55 hours, or 33 minutes. This is the same logarithm technique described earlier, just applied to a biology context. Once you’ve solved one exponential growth equation, you can solve them all.