\(y = a^x\) and \(y = \log_a x\) are inverses (reflections in \(y = x\)). The three log laws plus \(\log_a a = 1\) and \(\log_a 1 = 0\) let you solve \(a^x = b\) and equations that reduce to it; the change-of-base formula is not required, but \(\log_4 x = \frac12\log_2 x\) style relations are.
Exponentials (MM5.1, MM5.3)
- \(y = a^x\) (\(a > 1\)): through \((0, 1)\), increasing, asymptote \(y = 0\); for \(0 < a < 1\) it decreases. \(a^{-x} = \left(\frac1a\right)^x\).
- Compare growth by finding when curves cross: \(2^x\) and \(x^2\) meet at \(x = 2, 4\) and once for negative \(x\).
- \(a^x = b \Leftrightarrow x = \log_a b\). Equations like \(25^x - 3\cdot5^x + 2 = 0\) are quadratics in \(5^x\); keep only positive roots.
- Halving/doubling: \(\left(\frac12\right)^{t/5} = \frac{1}{10} \Rightarrow t = \frac{5}{\log_{10} 2}\).
Logarithms (MM5.2)
- \(\log_a(xy) = \log_a x + \log_a y\); \(\log_a\frac xy = \log_a x - \log_a y\); \(k\log_a x = \log_a x^k\); \(\log_a\frac1x = -\log_a x\); \(\log_a a = 1\), \(\log_a 1 = 0\).
- \(\log_a x\) is only defined for \(x > 0\): after solving \(\log_3(x + 6) + \log_3(x - 2) = 2\), reject any \(x\) that makes an argument non-positive.
- Chains: \(\log_a b = 2\) and \(\log_b c = 3\) give \(c = b^3 = a^6\), so \(\log_a c = 6\).
- \(\frac{\log 27}{\log 3} = 3\) because \(27 = 3^3\) (no change of base needed).
- Sums of logs telescope: \(\sum \log\frac{n+1}{n} = \log(N + 1)\).
Must-know (no formula booklet)
- \(\log_a(xy) = \log_a x + \log_a y\)
- \(k\log_a x = \log_a x^k\)
- \(a^x = b \Leftrightarrow x = \log_a b\)
- \(\log_a a = 1\), \(\log_a 1 = 0\)
- \(y = a^x\) and \(y = \log_a x\) are reflections in \(y = x\)
Classic traps
- \(\log(x + y)\) has no expansion; \(\log x - \log y\) is \(\log\frac xy\), not \(\frac{\log x}{\log y}\).
- Extraneous solutions: \(\log\) of a negative number is undefined even if the algebra 'works'.
- A translation of \(y = 2^x\) by \(3\) to the right is the same as a vertical stretch by \(\frac18\).
Worked example
Q. Solve \(2^{2x} - 3\cdot2^{x+2} + 32 = 0\).
A. With \(y = 2^x\): \(y^2 - 12y + 32 = (y - 4)(y - 8) = 0\), so \(x = 2\) or \(3\).