Logarithms & Inequalities
Turning powers into sums, and learning to reason with "greater than" — two tools that quietly run through all of calculus
- The logarithm as the inverse of exponentiation, with its domain and base conditions.
- The laws of logarithms, the change-of-base rule, and natural vs common logs.
- The shape of the log graph for \(a>1\) and \(0
- To solve logarithmic and exponential equations — and check domains.
- The wavy-curve (sign) method for rational and polynomial inequalities.
- Exponential, logarithmic and modulus inequalities, plus the AM–GM inequality.
What Is a Logarithm?
Exponentiation asks "what is \(a^y\)?" The logarithm reverses the question: "to what power must I raise \(a\) to get \(x\)?" That power is \(\log_a x\). Logarithms were invented to turn multiplication into addition — and that single act of taming exponents is why they appear everywhere from pH to decibels to the growth rates of calculus.
Valid only when the base \(a>0,\ a\neq1\) and the argument \(x>0\). Two immediate consequences: \(\log_a 1=0\), \(\log_a a=1\), and the cancellation identities \(a^{\log_a x}=x\) and \(\log_a(a^{x})=x\).
You can only take the log of a positive number, and the base must be positive and not 1. Every logarithmic equation or inequality therefore comes with a hidden domain — find it first, before solving, and discard any answer that violates it.
The Laws of Logarithms
Every law of logs is just an exponent law read backwards. Because \(a^m\cdot a^n=a^{m+n}\), multiplying numbers adds their logs — that is the whole idea.
| Law | Statement |
|---|---|
| Product | \(\log_a(mn)=\log_a m+\log_a n\) |
| Quotient | \(\log_a\!\dfrac{m}{n}=\log_a m-\log_a n\) |
| Power | \(\log_a(m^{p})=p\,\log_a m\) |
| Base of a power | \(\log_{a^{p}} m=\dfrac{1}{p}\log_a m\) |
| Reciprocal of base | \(\log_a b=\dfrac{1}{\log_b a}\) |
Change of Base
Your calculator only knows two bases — 10 and \(e\). The change-of-base rule lets you compute a log in any base from those, and is the key to comparing logs with different bases.
Pick any convenient base \(b\). With \(b=10\) you get the common log \(\log x\); with \(b=e\) the natural log \(\ln x\). A special case worth memorising: \(\log_a b\cdot\log_b c\cdot\log_c a=1\).
The Logarithmic Graph
The graph of \(y=\log_a x\) is the mirror image of \(y=a^x\) across the line \(y=x\). Its shape depends entirely on whether the base exceeds 1 — and that single fact governs every log inequality.
Logarithmic Equations
To solve an equation with logs, collapse both sides to a single log (or remove the logs entirely), solve the resulting algebraic equation, then verify against the domain. The verification step is not a formality — extraneous roots are routine.
Both require \(f>0\) (and \(g>0\)). After solving, substitute every candidate back to confirm each argument is positive and each base is valid.
Reasoning with Inequalities
An inequality is a comparison, and the rules for manipulating it are mostly the familiar ones — with one famous exception that traps the unwary.
| Operation | Effect |
|---|---|
| Add / subtract any \(c\) | direction unchanged |
| Multiply / divide by \(c>0\) | direction unchanged |
| Multiply / divide by \(c<0\) | direction reverses |
| Take reciprocals (same sign) | direction reverses |
When clearing a denominator like \(\dfrac{1}{x-2}>3\), do not multiply by \(x-2\) — its sign is unknown, so you cannot tell whether to flip. Instead move everything to one side, combine into a single rational expression, and use the wavy-curve method below.
The Wavy-Curve (Sign) Method
For a rational or polynomial inequality, factor it, mark the roots on a number line, and track the sign across each interval. A single factor changes sign at its root; a squared factor touches zero without changing sign.
For \(>0\) take the positive intervals; for \(<0\) the negative ones. Exclude roots of the denominator always; include roots of the numerator only for \(\ge\) or \(\le\).
Exponential & Logarithmic Inequalities
Both reduce to a single principle: whether the underlying function rises or falls. The base decides everything.
| Inequality | \(a>1\) | \(0 |
|---|---|---|
| \(a^{f}>a^{g}\) | \(f>g\) | \(f |
| \(\log_a f>\log_a g\) | \(f>g>0\) | \(0 |
Modulus (Absolute Value) Inequalities
The modulus \(|x|\) measures distance from 0, so modulus inequalities are statements about distance — and that picture makes them easy.
"Less than" gives a single band around 0; "greater than" gives two outward rays. For \(|f(x)|
When an inequality contains two or more modulus signs (e.g. \(|x-1|+|x-3|<4\)), split the number line at each critical point, drop the modulus signs with the correct sign on each piece, and solve each interval separately. Then union the partial solutions.
Strategies & Standard Results
A few results and habits that resolve most problems in this chapter.
For non-negatives, \(\dfrac{a+b}{2}\ge\sqrt{ab}\), with equality only when \(a=b\). The fastest route to many minimum/maximum bounds.
For any log expression, write down where every argument is positive before manipulating. Intersect at the end.
Set \(t=\log_a x\) or \(t=a^x\) to turn a tangled log/exponential equation into a polynomial in \(t\).
Putting It to Work
Problem. Evaluate \(\log_2 48-\log_2 3+\log_2 4\).
Solution. Combine with the product and quotient laws:
Problem. Show that \(\log_2 3\cdot\log_3 4\cdot\log_4 5\cdot\log_5 8=3\).
Solution. Write every log in base \(e\); the chain telescopes:
Problem. Solve \(\log_2(x-1)+\log_2(x+1)=3\).
Solution. Domain: \(x>1\). Combine and remove the log:
Only \(x=3\) satisfies the domain \(x>1\); reject \(x=-3\).
Problem. Solve \(\dfrac{(x-1)(x+2)}{x-3}\ge0\).
Solution. Roots at \(-2,1\) (numerator) and \(3\) (denominator, excluded). Signs alternate \(+\,-\,+\,-\) from the right… reading the positive/zero regions:
Endpoints \(-2,1\) are included (numerator zero, "\(\ge\)"); \(3\) is excluded.
Problem. Solve \(\log_{1/2}(x-1)>2\).
Solution. Base \(\tfrac12<1\), so the log is decreasing — the inequality flips when we remove it. Domain: \(x>1\).
Intersecting with \(x>1\) gives \(1
Problem. Solve \(|2x-3|\le5\).
Solution. The "less than" pattern gives a band:
So \(x\in[-1,\,4]\).
Chapter Summary
\(\log_a x=y\iff a^y=x\), with \(a>0,a\neq1,x>0\). Logs invert powers.
Product adds, quotient subtracts, power multiplies; change base by \(\log_a x=\tfrac{\log_b x}{\log_b a}\).
\(a>1\) keeps the inequality; \(0
Factor, mark roots, alternate signs; exclude denominator roots, mind squared factors.
\(|x|
\(\tfrac{a+b}{2}\ge\sqrt{ab}\), equality at \(a=b\) — the workhorse for bounds.
Problems
For each, find the domain before solving. Difficulty rises down the list.
- Evaluate \(\log_3 81+\log_5 \tfrac{1}{25}-\log_2 \sqrt{8}\).
- If \(\log_{10}2=0.3010\), find the number of digits in \(2^{50}\).
- Solve \(\log_3(x+2)+\log_3(x-2)=\log_3 5\)… (state why one root is rejected).
- Solve \(2^{2x}-5\cdot2^{x}+4=0\) by substituting \(t=2^x\).
- Solve \(\dfrac{x-1}{x+2}<2\) using the wavy-curve method.
- Solve \(\log_{0.5}(x^2-5x+6)\ge -1\).
- Solve \(|x-2|+|x-5|\le 9\).
- Solve \(3^{x+1}>5^{x-1}\), leaving the answer in terms of logs.
- Find the minimum value of \(x+\dfrac{4}{x}\) for \(x>0\) using AM–GM.
- Solve \(\log_x 2\cdot\log_{2x}2=\log_{4x}2\).
- Solve the inequality \(\dfrac{x^2-3x+2}{x^2-1}\ge0\), taking care with repeated and common factors.
- Find all \(x\) satisfying \(\log_2(\log_3(\log_4 x))=0\).