Everything from the 4 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.
\(\log_a x = y \iff a^y = x\) (for \(a > 0\), \(a \ne 1\), \(x > 0\)). The laws: \(\log_a xy = \log_a x + \log_a y\), \(\log_a \dfrac{x}{y} = \log_a x - \log_a y\), \(\log_a x^k = k\log_a x\). Always check that every logarithm in the original equation is of a positive number.
\(\log_2 \dfrac{1}{8}\) is \(-3\), not \(\dfrac{1}{3}\) or \(-\dfrac{1}{3}\). Negative logs come from fractions; fractional logs come from roots.
There is no law for \(\log(x + y)\): \(\log_a(x + y) \ne \log_a x + \log_a y\). Also \(\dfrac{\log_a x}{\log_a y}\) is NOT \(\log_a \dfrac{x}{y}\).
Do not 'cancel the logs' term by term: \(\log_2(x + 3) + \log_2(x - 1) = 5\) does NOT give \((x + 3) + (x - 1) = 5\) or \(= 32\). Make ONE log first, then undo it.
\(\ln x = \log_e x\), so \(y = e^x \iff x = \ln y\). They undo each other: \(\ln(e^x) = x\) for all \(x\), and \(e^{\ln x} = x\) for \(x > 0\). \(y = e^{kx}\) always passes through \((0, 1)\), stays above the \(x\)-axis, grows for \(k > 0\) and decays towards \(y = 0\) for \(k < 0\).
\(e^{-x}\) is never negative: \(y = e^{-x}\) is a decreasing curve ABOVE the \(x\)-axis, not the curve \(y = e^x\) flipped below it. The minus sign in the power reflects in the \(y\)-axis; a minus sign in front, \(-e^x\), reflects in the \(x\)-axis.
\(\ln(e^x + 1)\) is NOT \(x + 1\), and \(\ln(2e^x)\) is \(\ln 2 + x\), not \(2x\). ln only undoes \(e\) when the whole inside of the log is a single power of \(e\).
Take ln only after isolating the exponential: from \(3e^{x} + 1 = 10\), first get \(e^x = 3\), then \(x = \ln 3\). Writing \(\ln 3 + x + \ln 1 = \ln 10\) is not a valid step, because there is no law for the log of a sum.
Take ln (or lg) of both sides: \(a^{f(x)} = b \Rightarrow f(x)\ln a = \ln b\). For an inequality, dividing by \(\ln a\) keeps the sign when \(a > 1\) but REVERSES it when \(0 < a < 1\), because then \(\ln a < 0\).
\(\dfrac{\ln 5}{\ln 2}\) is NOT \(\ln \dfrac{5}{2}\) and not \(\ln 5 - \ln 2\). A quotient of logs does not simplify: work it out as one number divided by another.
In \(n\ln 0.8 < \ln 0.1\), both logs are negative. Forgetting to reverse gives \(n < 10.3\), which says the SMALL values work, the opposite of what happens as \(0.8^n\) shrinks.
\(2^{2x + 1}\) is \(2 \times 2^{2x} = 2u^2\), not \(u^2 + 1\) and not \(2u\). Split the index with \(a^{m + n} = a^m \times a^n\) before you substitute.
\(y = kx^n \Rightarrow \ln y = \ln k + n\ln x\): plot \(\ln y\) against \(\ln x\); gradient \(n\), intercept \(\ln k\). \(y = ka^x \Rightarrow \ln y = \ln k + x\ln a\): plot \(\ln y\) against \(x\); gradient \(\ln a\), intercept \(\ln k\). Compare with \(Y = mX + c\).
For \(y = ka^x\) the horizontal axis is plain \(x\), not \(\ln x\): the \(x\) is already in the index, and taking logs brings it down as it is. Plot \(\ln x\) only when \(x\) is raised to a constant power.
The coordinates in these questions are values of \(\ln y\) (and perhaps \(\ln x\)), not of \(y\). The intercept of the line is \(\ln k\); answering \(k = 2.6\) instead of \(k = e^{2.6}\) loses the final marks.
Undo with the SAME base you took logs in. If the axis is \(\lg y\), use \(10^{1.2}\); using \(e^{1.2}\) gives \(3.32\), a completely different (wrong) constant.