Using logarithms to solve exponential equations. Graphs of exponential functions.
Answer at least 3 of 3 correctly to complete this section.
Exponential equations appear everywhere: radioactive decay, compound interest, population growth, cooling curves. In every case, the unknown is in the exponent — and you cannot solve that with algebra alone. You need logarithms.
This lesson builds directly on the Laws of Logarithms. If you are not confident with the three log laws and converting between exponential and logarithmic form, review that lesson first.
The graph of y=ax (where a>0, a=1) has a distinctive shape that you must recognise in exams.
| Feature | Value |
|---|---|
| y-intercept | Always (0,1) because a0=1 |
| Asymptote | The x-axis (y=0) — the curve approaches but never touches it |
| Domain | All real numbers |
| Range | y>0 |
| Shape (a>1) | Increasing — growth |
| Shape (0<a<1) | Decreasing — decay |
y=3⋅2xstretches y=2x vertically by factor 3
y=2x+1translates y=2x one unit left
y=2x+5translates y=2x five units up (asymptote becomes y=5)
Exam tip: When a graph has a horizontal asymptote at y=k (not y=0), the equation is of the form y=ax+k. Read k from the graph before solving.
Answer at least 3 of 4 correctly to complete this section.
When both sides of an equation can be written as powers of the same base, you can solve without logarithms.
If af(x)=ag(x), then f(x)=g(x)
This works because exponential functions are one-to-one: each input gives a unique output.
Solve 2x=32.
Rewrite the right-hand side as a power of 2:
2x=25
Since the bases are equal:
x=5
Solve 32x+1=81.
Rewrite 81 as a power of 3: 81=34.
32x+1=34
Equate the exponents:
2x+1=4
2x=3
x=23
Solve 4x=8.
Both 4 and 8 are powers of 2: 4=22 and 8=23.
(22)x=23
22x=23
2x=3⟹x=23
When to use this method: If you can spot that the right-hand side is an integer power of the base (or both sides share a common base), matching bases is faster and gives an exact answer without a calculator.
Answer at least 3 of 4 correctly to complete this section.
When you cannot match bases, take the logarithm of both sides and use the power law to bring the exponent down.
Starting with af(x)=b:
You can use log10, ln, or any base — the method works with all of them.
Solve 3x=20, giving your answer to 3 significant figures.
log(3x)=log(20)
xlog(3)=log(20)
x=log(3)log(20)=0.47711.301=2.73 (3 s.f.)
Solve e3x=20.
Since the base is e, use ln (the natural logarithm):
ln(e3x)=ln(20)
3x=ln(20)(since ln(e)=1)
x=3ln(20)≈0.999
For an exact answer: x=3ln20.
Solve 52x−1=40.
log(52x−1)=log(40)
(2x−1)log(5)=log(40)
2x−1=log(5)log(40)=0.6991.602=2.292
2x=3.292
x=1.65 (3 s.f.)
Choose an equation below and work through each step:
Choose an equation and step through the solution
Common mistake: Students take log of only one side. You must apply the logarithm to both sides of the equation — it is a function applied to both sides, just like dividing both sides by the same number.
Answer at least 3 of 4 correctly to complete this section.
Solve 2x+3=19.
First, isolate the exponential:
2x=16
Now 16=24, so x=4.
Key step: Always isolate the exponential term before taking logs or matching bases.
The value of a car is modelled by V=15000×0.85t, where V is the value in pounds and t is the time in years. After how many years will the car be worth less than £5000?
15000×0.85t<5000
0.85t<31
log(0.85t)<log(31)
tlog(0.85)<log(31)
Since log(0.85)<0, flip the inequality when dividing:
t>log(0.85)log(1/3)=−0.1625−0.4771=6.76
So after 7 complete years, the car is worth less than £5000.
Watch out: When you divide by a negative number (like log(0.85)), the inequality reverses. This catches many students out in modelling questions.
Solve 3e2x=15.
e2x=5
2x=ln5
x=2ln5≈0.805 (3 s.f.)
| Pattern | Technique |
|---|---|
| ax=an | Match bases, equate exponents |
| ax=b (can’t match) | Take log of both sides, use power law |
| ef(x)=k | Take ln of both sides |
| kax=m | Divide by k first, then solve |
| ax+c=d | Subtract c first to isolate the exponential |
Answer at least 3 of 4 correctly to complete this section.
Lock in what you've learned with exam-style questions and spaced repetition.