Unit 6 · Index and Logarithm Laws
Define a logarithm as an index, use the product, quotient and power laws, and solve logarithmic equations while rejecting invalid roots.
Since $2^3 = 8$, we say "the logarithm of $8$ to base $2$ is $3$", written $\log_2 8 = 3$. In general $$\log_a x = y \iff a^y = x \qquad (a > 0,\ a \ne 1,\ x > 0).$$ So a logarithm answers the question "what index must the base carry to give $x$?". For example $\log_3 81 = 4$ because $3^4 = 81$; $\log_2 \dfrac18 = -3$ because $2^{-3} = \dfrac18$; $\log_{25} 5 = \dfrac12$ because $25^{1/2} = 5$.
Immediate consequences of the definition: $$\log_a 1 = 0, \qquad \log_a a = 1, \qquad a^{\log_a x} = x, \qquad \log_a a^y = y.$$ $x$ must be positive: no power of a positive base is zero or negative. The logarithm to base $10$ is the common logarithm (පොදු ලඝුගණකය), written $\log x$.
Let $\log_a x = p$ and $\log_a y = q$, so $x = a^p$ and $y = a^q$. Then $xy = a^{p+q}$, which says $\log_a (xy) = p + q$. The other laws follow in the same way from the laws of indices.
$$\log_a (xy) = \log_a x + \log_a y$$ $$\log_a \frac{x}{y} = \log_a x - \log_a y$$ $$\log_a x^k = k \log_a x$$
Example: Evaluate $\log_6 4 + \log_6 9$. By the product law this is $\log_6 36 = 2$, since $6^2 = 36$.
Example: Given $\log 2 = a$ and $\log 3 = b$ (base $10$), write $\log 24$ in terms of $a$ and $b$. $24 = 2^3 \times 3$, so $\log 24 = 3\log 2 + \log 3 = 3a + b$.
Combine the logarithms into one using the laws, change to the index form, solve, then check. The arguments of the original logarithms must be positive.
Example: Solve $\log_3 (x + 2) + \log_3 x = 1$. Product law: $\log_3 [x(x+2)] = 1$, so $x(x+2) = 3^1 = 3$, $x^2 + 2x - 3 = 0$, $(x+3)(x-1) = 0$. $x = -3$ is rejected because $\log_3 (-3)$ is undefined. Hence $x = 1$; check: $\log_3 3 + \log_3 1 = 1 + 0 = 1$.
Example: Solve $\log_2 x - \log_2 (x - 1) = 2$. Quotient law: $\log_2 \dfrac{x}{x-1} = 2$, so $\dfrac{x}{x-1} = 4$, $x = 4x - 4$, $x = \dfrac43$. Both $x$ and $x - 1 = \dfrac13$ are positive, so the root is valid.
Exam tip: A Paper I logarithm question usually ends with "show that $x = \dots$" or "find the value of $x$". Always state the condition (such as $x > 1$) at the start and reject the root that violates it; examiners award a mark for this. Do not write $\log(x + y) = \log x + \log y$: only the product has a sum law.