MISSION · Algebra
Logarithms
The question "ten to the what?" written as an operation
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The idea in one sentence
A logarithm is an exponent — specifically, the exponent you’d have to raise the base to in order to hit a given number.
\[\log_{10} 1000 = 3 \quad \text{because} \quad 10^3 = 1000\]Reading it aloud helps
\(\log_b a = c\) is read “the log, base \(b\), of \(a\)”, and it means “\(b\) to the power \(c\) equals \(a\).” Those two statements are the same fact written twice:
\[\log_2 32 = 5 \quad \Longleftrightarrow \quad 2^5 = 32\]Anything confusing about logs usually dissolves by translating back to exponent form. Why is \(\log_b 1 = 0\) for every base? Because \(b^0 = 1\) always. Why is the log of a negative number undefined? Because no exponent turns a positive base into a negative result.
Logs and exponentials are inverse operations, in the same way squaring and square-rooting are. One undoes the other, which is exactly why logs are the tool for solving equations where the unknown is stuck in an exponent.
The laws, and why they hold
\(\log(xy) = \log x + \log y\) \(\log\left(\frac{x}{y}\right) = \log x - \log y\) \(\log(x^n) = n \log x\)
These look arbitrary until you remember that a log is an exponent. Multiplying powers of the same base adds their exponents — so the log of a product is the sum of the logs. Every law here is an exponent law, translated.
That first law was the original point of logarithms. Before calculators, multiplying two seven-digit numbers meant looking up two logs, adding them, and looking the answer back up. Logs turned multiplication into addition, and they did the heavy arithmetic of navigation and astronomy for three hundred years.
Why they’re still everywhere
Because logs compress. A scale that runs from 1 to 1,000,000 is unusable; its logarithm runs from 0 to 6.
- Decibels — sound intensity spans a trillion-to-one range from a whisper to a jet engine.
- The Richter scale — each whole step is ten times the amplitude, which is why a magnitude 7 is not slightly worse than a 6.
- pH — the concentration of hydrogen ions spans fourteen orders of magnitude.
In each case equal steps on the scale mean equal multiplications in the quantity. That’s what a log scale is for, and misreading one as linear badly understates how big the big end is.
Solving with them
\(2^x = 50\) can’t be solved by ordinary rearranging — the unknown is an exponent. Take logs of both sides and the third law brings it down:
\[x \log 2 = \log 50 \quad \Rightarrow \quad x = \frac{\log 50}{\log 2} \approx 5.64\]Common mix-ups
- Splitting the log of a sum. \(\log(x + y)\) is not \(\log x + \log y\). The laws cover products, not sums.
- Reading \(\log x^2\) as \((\log x)^2\). The first is \(2\log x\); the second is the log squared.
- Forgetting the base. \(\log\) usually means base 10, \(\ln\) means base \(e\), and in computing \(\log\) often means base 2. Check which.
- Expecting a log of a negative number. There isn’t one.
Check yourself
Evaluate \(\log_3 81\), and solve \(5^x = 200\).
\(\log_3 81 = 4\), since \(3^4 = 81\). For the second, \(x = \frac{\log 200}{\log 5} \approx 3.29\).
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