Exponential inequalities: look at the base
An exponential inequality has the unknown in the power, like 2ˣ > 8. (A power is the small raised number.)
Step 1: write both sides with the same base: 2ˣ > 2³.
Step 2: compare the powers. The base decides:
- Base above 1 (like 2 or 10): the curve rises. The sign stays. 2ˣ > 2³ gives x > 3.
- Base between 0 and 1 (like ½): the curve falls. The sign flips. (½)ˣ > (½)³ gives x < 3.
A power is never zero or negative, so something like 2ˣ > −5 is true for every x, and 2ˣ < −5 is never true.
Hidden quadratic: for 4ˣ − 3·2ˣ + 2 < 0, put t = 2ˣ (t > 0). Then t² − 3t + 2 < 0, so 1 < t < 2, so 0 < x < 1.
Logarithmic inequalities: domain first
A logarithm answers "which power gives this number?". log₂ 8 = 3 because 2³ = 8. It only exists when the number inside is greater than 0.
To solve loga f(x) < loga g(x):
- Write the domain: f(x) > 0 and g(x) > 0.
- Base above 1: keep the sign, f < g. Base between 0 and 1: flip it, f > g.
- Join the answer with the domain (take only the part that is in both).
Example: log₂(x − 1) < 3. Domain: x > 1. Compare: x − 1 < 8, so x < 9. Answer: 1 < x < 9.
Turn a plain number into a log when needed: 3 = log₂ 8.
Irrational (root) inequalities
An irrational inequality has the unknown under a root, like √(x + 2) < x. A square root is never negative, and the thing under it cannot be negative.
Case √f < g (root is small). All three must hold together: f ≥ 0, g > 0, and f < g².
Case √f > g (root is big). Two roads, then join them with "or":
- g < 0 and f ≥ 0 (a root that is 0 or more always beats a negative number), or
- g ≥ 0 and f > g².
Never square both sides unless you know both sides are not negative. Squaring is only safe then.
Example: √(x + 2) < x. Domain x ≥ −2. Need x > 0. Square: x + 2 < x², so x² − x − 2 > 0, so x > 2 or x < −1. With x > 0 the answer is x > 2.
Check every answer
Pick one number inside your answer and one outside. Put both in the original inequality. Inside must say TRUE, outside must say FALSE (or "not allowed"). In the 3D board the green bar is the true zone and the red bar is the forbidden zone.
For mixed problems use a sign chart: mark the points where each part changes, test a number in every gap, and keep the gaps that are true.
Try it: predict, then check
In the 3D board pick 2^x > 4. Before you slide, guess: is x = 1 in the green bar? Slide the test x to 1 and read TRUE or FALSE. Now pick (1/2)^x > 4 and guess again for x = 1. At home: fold a paper in half again and again. After 5 folds there are 2⁵ = 32 layers. How many folds give more than 1000 layers? (2ˣ > 1000 gives 10 folds.)
Key formulas and definitions
- a > 1: aᶠ > aᵍ ⇔ f > g
- 0 < a < 1: aᶠ > aᵍ ⇔ f < g
- a > 1: logₐ f > logₐ g ⇔ f > g > 0
- 0 < a < 1: logₐ f > logₐ g ⇔ 0 < f < g
- √f < g ⇔ f ≥ 0, g > 0, f < g²
- √f > g ⇔ (g < 0, f ≥ 0) or (g ≥ 0, f > g²)
Worked examples
1. Solve 3ˣ > 27.
27 = 3³, so 3ˣ > 3³. Base 3 is above 1, so keep the sign: x > 3.
2. Solve (1/2)^(x − 1) ≥ 1/8.
1/8 = (1/2)³. So (1/2)^(x − 1) ≥ (1/2)³. Base is between 0 and 1, so flip: x − 1 ≤ 3, so x ≤ 4.
3. Solve log₂(x − 1) < 3.
Domain: x − 1 > 0, so x > 1. Since 3 = log₂ 8, we need x − 1 < 8, so x < 9. Answer: 1 < x < 9.
4. Solve log_{1/3}(2x − 1) > −1.
Domain: 2x − 1 > 0, so x > 1/2. Write −1 = log_{1/3} 3. Base is below 1, so flip: 2x − 1 < 3, so x < 2. Answer: 1/2 < x < 2.
5. Solve 4ˣ − 3·2ˣ + 2 < 0.
Let t = 2ˣ, t > 0. Then t² − 3t + 2 < 0, so (t − 1)(t − 2) < 0, so 1 < t < 2. That is 2⁰ < 2ˣ < 2¹, so 0 < x < 1.
6. Solve √(x + 2) < x.
Need x + 2 ≥ 0 (x ≥ −2), x > 0, and x + 2 < x². The last gives x² − x − 2 > 0, so (x − 2)(x + 1) > 0, so x > 2 or x < −1. Join with x > 0: x > 2.
7. Solve √(x + 2) > x.
Road 1: x < 0 and x + 2 ≥ 0, so −2 ≤ x < 0. Road 2: x ≥ 0 and x + 2 > x², so x² − x − 2 < 0, so −1 < x < 2, so 0 ≤ x < 2. Join: −2 ≤ x < 2.
8. Solve log₂ x + log₂(x − 2) ≤ 3.
Domain: x > 0 and x − 2 > 0, so x > 2. Combine: log₂(x(x − 2)) ≤ 3, so x² − 2x ≤ 8, so x² − 2x − 8 ≤ 0, so (x − 4)(x + 2) ≤ 0, so −2 ≤ x ≤ 4. With the domain: 2 < x ≤ 4.
Common mistakes
- Forgetting to flip the sign when the base is between 0 and 1. Always glance at the base first.
- Skipping the domain of a log. log(x − 1) needs x > 1, even if the algebra gives a wider answer.
- Squaring both sides of √f > g when g is negative. A negative g is already smaller than any root.
- Writing 2ˣ = t and forgetting t > 0. A power of 2 is never zero or negative, so throw away t ≤ 0.