TheIBAxolotl
Fractional exponents
The index laws force a meaning on fractional powers. Since \((a^{1/2})^2 = a^1\), the power \(\frac12\) must be the square root. In general
\[a^{\frac{1}{n}} = \sqrt[n]{a}, \qquad a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^m.\]
Take the root first to keep numbers small: \(8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4\), and \(27^{-1/3} = \frac{1}{3}\).
The laws of logarithms
Because logs are exponents, each index law has a log twin:
- Product: \(\log_a(xy) = \log_a x + \log_a y\)
- Quotient: \(\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y\)
- Power: \(\log_a(x^k) = k \log_a x\)
The power law is the most useful: it brings an unknown exponent down to ground level, where ordinary algebra can reach it.
Change of base. To evaluate a log in an awkward base, use \(\log_a x = \dfrac{\log_b x}{\log_b a}\) with any convenient \(b\), typically 10 or \(e\). For example, \(\log_2 7 = \frac{\ln 7}{\ln 2} \approx 2.81\).
Worked example: exponential equation
Solve \(3^{2x-1} = 5^{x}\).
Take natural logs of both sides and use the power law:
\[(2x - 1)\ln 3 = x \ln 5.\]
Collect the \(x\) terms: \(x(2\ln 3 - \ln 5) = \ln 3\), so
\[x = \frac{\ln 3}{\ln 9 - \ln 5} = \frac{\ln 3}{\ln \frac{9}{5}} \approx 1.87.\]
Worked example: log equation
Solve \(\log_2 x + \log_2(x - 2) = 3\).
Combine with the product law: \(\log_2\big(x(x-2)\big) = 3\), so \(x(x - 2) = 2^3 = 8\). Then \(x^2 - 2x - 8 = 0\) gives \((x - 4)(x + 2) = 0\). Now check the domain: \(x = -2\) would need the log of a negative number, so the only solution is \(x = 4\).
Simplifying
To write \(2\log 5 + \log 4 - \log 10\) as a single number: \(\log 25 + \log 4 - \log 10 = \log\frac{100}{10} = \log 10 = 1\). Always apply the power law first, then combine sums and differences.
Common mistakes
- \(\log(x + y) \ne \log x + \log y\). There is no law for the log of a sum.
- \(\frac{\log x}{\log y} \ne \log x - \log y\); the quotient law is about \(\log\frac{x}{y}\).
- \((\log x)^2 \ne 2\log x\); only \(\log(x^2) = 2\log x\).
- Forgetting to reject solutions that make an argument zero or negative.
- Dropping logs from a sum: \(\log x + \log 2 = \log 6\) gives \(2x = 6\) only after combining into \(\log 2x\) first.
Exam tips
- Non-calculator questions usually want an exact answer such as \(\frac{\ln 3}{\ln 1.8}\); do not round unless asked.
- If both sides can be written as powers of the same base, equate the exponents directly; it is faster than logs.
- Show the line where you combine logs into one; it is the key method step.