Exponents and logarithms
A logarithm is an exponent — that one sentence explains every law on this page. Here is the translation between the two forms, the five laws with the reason each one works, a decision tree for solving equations, and the growth and decay setup that word problems keep reusing.
- 1. A logarithm is an exponent
- 2. The laws, and why
- 3. Solving: which method?
- 4. Growth and decay
- 5. The two graphs
- 6. Traps
A logarithm is an exponent
The idea: logb c = a and ba = c say exactly the same thing. The logarithm answers "what exponent turns b into c?"
| Logarithmic form | Exponential form | Said out loud |
|---|---|---|
| log₂ 8 = 3 | 2³ = 8 | 2 to the what gives 8? Three. |
| log₅ 1 = 0 | 5⁰ = 1 | Any base to the zero is 1. |
| log₃(1/9) = −2 | 3⁻² = 1/9 | Negative exponents give fractions. |
| log₉ 3 = 1/2 | 9^(1/2) = 3 | A fractional exponent is a root. |
- log x with no base written means base 10. ln x means base e.
- The argument must be positive. log(x − 3) only exists for x > 3, which is where restrictions on solutions come from.
- The base is positive and not 1, because 1 raised to anything is always 1.
Translating is usually the whole question
Solve log₄ x = 3/2. Rewrite it as 4^(3/2) = x, which is (√4)³ = 2³ = 8. No laws needed — just the translation.
The laws, and why
The idea: the log laws are the exponent laws wearing a different hat. Multiplying powers adds exponents, so multiplying arguments adds logarithms.
| Law | Rule | Because, in exponent form |
|---|---|---|
| Product | logb(MN) = logb M + logb N | bx · by = bx+y |
| Quotient | logb(M/N) = logb M − logb N | bx ÷ by = bx−y |
| Power | logb(Mⁿ) = n logb M | (bx)n = bxn |
| Change of base | logb c = loga c / loga b | lets your calculator do any base |
| Inverses | logb(bx) = x and b^(logb x) = x | the two functions undo each other |
Worked: write 2 log x − ½ log y + log 3 as a single logarithm
- Power law first, to clear every coefficient: log x² − log √y + log 3.
- Products and quotients: addition puts a factor on top, subtraction puts it on the bottom.
- Answer: log(3x² / √y).
Going the other way — expanding log(3x²/√y) into separate logs — is the same steps in reverse, and is just as commonly asked.
Worked: logb 50 in terms of m and n, given logb 2 = m and logb 5 = n
Factor the argument into the pieces you were given: 50 = 2 × 5². So logb 50 = logb 2 + 2 logb 5 = m + 2n. These questions are always "factor the number, then apply the laws".
Solving: which method?
The idea: three questions decide the method. Can both sides be written with the same base? Is the variable in an exponent? Is there more than one logarithm?
| If the equation… | Do this | Example |
|---|---|---|
| can be written with a common base | equate the exponents | 4^(x+1) = 8^(x−2) → 2^(2x+2) = 2^(3x−6) → x = 8 |
| has the variable in an exponent, no common base | take the log of both sides, then use the power law | 3^x = 20 → x log 3 = log 20 → x ≈ 2.727 |
| has one logarithm | rewrite in exponential form | log₂(x − 1) = 4 → x − 1 = 16 → x = 17 |
| has two or more logarithms, same base | combine into one with the laws, then rewrite | log x + log(x − 3) = 1 → log(x² − 3x) = 1 → x² − 3x = 10 |
That last example finishes as x² − 3x − 10 = 0, so x = 5 or x = −2. But log(−2) does not exist, so x = 5 is the only solution. Any time you solve a logarithmic equation, check every answer in the original equation — rejecting an extraneous root is usually worth a mark on its own.
Watch out: log 8 / log 2 is not log 4. It is the change-of-base form of log₂ 8, which is 3. Dividing logs is a division of two numbers, not a quotient law.
Growth and decay
The idea: one formula, y = a·b^(t/p), covers doubling, halving, interest and depreciation. Read the question for the four letters.
| Letter | Means | Typical wording |
|---|---|---|
| a | the starting amount | "initially", "a 250 mg dose" |
| b | what it multiplies by each period | doubles → 2; halves → ½; grows 6% → 1.06; falls 15% → 0.85 |
| p | how long one period lasts | "half-life of 5.7 days", "doubles every 3 hours" |
| t | elapsed time, in the same units as p | what you solve for, or substitute |
Worked: a 100 mg sample with a half-life of 5.7 days — when is 12 mg left?
- Model: A = 100(½)^(t/5.7).
- Substitute: 12 = 100(½)^(t/5.7), so (½)^(t/5.7) = 0.12.
- Take logs: (t/5.7)·log 0.5 = log 0.12.
- Solve: t = 5.7 · log 0.12 / log 0.5 = 17.435… ≈ 17.4 days.
Round only at the end, and sanity-check: three half-lives is 17.1 days and would leave 12.5 mg, so 17.4 days for 12 mg is right where it should be.
- Percent growth: "increases 6% per year" makes b = 1.06 with p = 1. "Decreases 15%" makes b = 0.85 — subtract from 1, do not use −0.15.
- Different periods: "doubles every 3 h" is b = 2, p = 3. If t is in minutes, p must be 180.
- Solving for b: if a quantity grows from 500 to 1 800 in 4 years, then 1 800 = 500·b⁴, so b⁴ = 3.6 and b = 3.6^(1/4) ≈ 1.377 — about 37.7% growth per year.
The two graphs
The idea: y = bx and y = logb x are reflections of each other in the line y = x, so every feature of one is the mirror image of the other.
| y = bx (b > 1) | y = logb x (b > 1) | |
|---|---|---|
| Domain | all real numbers | x > 0 |
| Range | y > 0 | all real numbers |
| Asymptote | horizontal, y = 0 | vertical, x = 0 |
| Passes through | (0, 1) | (1, 0) |
Transformations then behave exactly as they do for any function. In y = log₂(x + 4) − 3, the +4 inside moves the graph 4 left, so the asymptote moves from x = 0 to x = −4 and the domain becomes x > −4. The −3 slides it down and changes nothing about the domain.
Traps
The idea: almost every lost mark here is one of five things.
- log(a + b) is not log a + log b. The product law applies to a product inside one log, never to a sum.
- Coefficients must be cleared first. 2 log x is log x², not log 2x.
- Extraneous roots. Any solution that makes an argument zero or negative is rejected.
- Same base or nothing. Laws only combine logs that share a base; otherwise change base first.
- Percent decrease. A 15% drop is b = 0.85. Writing b = 0.15 models keeping 15%, which is a much faster decay.