Solving Exponential Equations with Logarithms
Take logarithms of both sides to bring an unknown down from an exponent, and solve equations that are quadratic in eˣ by substitution.
What a learner can do afterwards
- Solve 3ˣ = 20 to three significant figures
- Solve e^(2x) - 5eˣ + 6 = 0 using a substitution
- Find how long a modelled quantity takes to halve
1 · Read
You isolate the exponential part first, then take logs of both sides. The power rule pulls the exponent down front, and that log step is what frees x. Take 3^x = 20: logs give x log 3 = log 20, so x = log 20/log 3, about 2.73. Round only at the very end.
When both sides share a base, skip the logs and match the exponents. Since 125 is 5^3, the equation 5^x = 125 gives x = 3 at once. Use this shortcut whenever the same base appears on both sides.
Treat e^(2x) - 5e^x + 6 = 0 as a quadratic in disguise. Substitute y = e^x to get y^2 - 5y + 6 = 0, which factors to y = 2 or y = 3. Undo the swap with logs: x = ln 2 or x = ln 3, not 2 or 3.
For halving stories, count how many half lives fit and halve that many times. A sample of 80 g with a 4 hour half life gives 3 half lives in 12 hours, so 80 to 40 to 20 to 10 g. A tracer falling from 200 to 25 MBq halves 3 times, and at 6 hours each that takes 18 hours.
Free x with logs, swap e^x for y in quadratic cases, and count half lives in decay stories.
2 · Watch
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8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.