A logarithm is an exponent
Tonight's idea
\(\log_2 8\) is a question: what power of 2 gives 8? The answer is 3. That is the whole definition:
\(\log_b x = y\) means exactly \(b^y = x\). A logarithm is an exponent.
So \(\log_{10} 1000 = 3\), \(\log_5 \tfrac{1}{25} = -2\), \(\log_9 3 = \tfrac12\), and \(\log_b 1 = 0\) for every base. Plain \(\log\) means base 10; \(\ln\) means base \(e \approx 2.718\), met properly in The number e.
Every log law is an exponent law read backwards. Let \(x = b^m\) and \(y = b^n\), so \(m = \log_b x\) and \(n = \log_b y\). Then \(xy = b^{m+n}\), so \(\log_b(xy) = \log_b x + \log_b y\): multiplying numbers adds their logs. Similarly \(\log_b(x/y) = \log_b x - \log_b y\), and \(x^k = b^{mk}\) gives \(\log_b(x^k) = k\log_b x\): powers come down as multipliers, which is what solves equations.
Change of base. To get \(\log_2 10\) on a calculator: if \(2^y = 10\) then \(y \log 2 = \log 10 = 1\), so \(y = 1/\log 2 \approx 3.32\). In general \(\log_b x = \dfrac{\log x}{\log b}\).
Counting digits. \(N\) has \(\lfloor \log_{10} N \rfloor + 1\) digits, where \(\lfloor\;\rfloor\) means round down to a whole number. For \(2^{1000}\): \(1000 \log_{10} 2 \approx 301.03\), so \(2^{1000}\) has 302 digits.
Watch one: two equations
\(5 \cdot 3^x = 200\). Isolate the power: \(3^x = 40\). Take logs: \(x \ln 3 = \ln 40\), so \(x = \dfrac{\ln 40}{\ln 3} \approx 3.358\). (Sanity: \(3^3 = 27\), \(3^4 = 81\).)
\(\log_2 x + \log_2(x - 2) = 3\). Combine: \(\log_2\big(x(x-2)\big) = 3\), so \(x(x-2) = 8\), so \(x = 4\) or \(x = -2\). Reject \(-2\): no log of a negative. Answer \(x = 4\). Always check log equations against the domain. If \(x\) ends up on both sides after taking logs, collect the \(x\) terms and factor \(x\) out.
Your turn
- Evaluate without a calculator: \(\log_2 64\), \(\;\log_3 \tfrac{1}{27}\), \(\;\log_8 4\), \(\;\ln e^3\).
- Solve to three decimal places: \(7^x = 50\); \(\;2^{x+1} = 3^x\).
- How many digits are in \(3^{500}\)? Use \(\log_{10} 3 \approx 0.47712\).
- Money grows at 6% per year. How many years until it doubles? Bankers use the "rule of 72": doubling time \(\approx 72/(\text{percent rate})\). Check the rule at 6% and 12% against the exact answers.
Stretch: a log you can prove is irrational
Suppose \(\log_2 3 = p/q\) with positive whole numbers \(p, q\). Then \(2^{p/q} = 3\), so \(2^p = 3^q\). Why is that impossible? Conclude that \(\log_2 3\) is irrational, then adapt the argument to \(\log_{10} 2\).
Answers
- \(6\); \(\;-3\); \(\;\tfrac23\) (since \(8^{2/3} = 4\)); \(\;3\).
- \(x = \dfrac{\ln 50}{\ln 7} \approx 2.010\). \(\;(x+1)\ln 2 = x\ln 3\), so \(x = \dfrac{\ln 2}{\ln 3 - \ln 2} = \dfrac{\ln 2}{\ln 1.5} \approx 1.710\).
- \(500 \times 0.47712 = 238.56\), so \(239\) digits.
- \(1.06^n = 2\), so \(n = \dfrac{\ln 2}{\ln 1.06} \approx 11.9\) years; the rule says 12. At 12%: exact 6.1, rule says 6. It works because \(\ln 2 \approx 0.693\) and \(\ln(1 + r) \approx r\) for small \(r\), and 72 is a nearby number with many divisors.
Stretch. \(2^p\) is even and \(3^q\) is odd, so they cannot be equal; no such fraction exists. For \(\log_{10} 2 = p/q\): \(10^p = 2^q\), but the left side is divisible by 5 and the right side is not. So the decimal \(0.30102999\ldots\) never repeats.