6.3 Logarithmic functions
A logarithm answers the question “what power do I need?”.
Definition 6.7. For \(a>0\), \(a\neq 1\) and \(x>0\), \[\log _a x=y\quad \text {means exactly}\quad a^y=x.\]
The two statements say the same thing in different words, and being able to move between them is most of the skill. For instance \(\log _2 8=3\) because \(2^3=8\).
Note 6.8. The logarithm of a negative number or of zero does not exist, because \(a^y\) is always positive. This restriction is what forces the rejection of certain roots later in this section, and it is not optional.
Two bases are used so often they have their own notation: \(\log x\) means \(\log _{10}x\), and \(\ln x\) means \(\log _e x\), where \(e\approx 2.718\).
6.3.1 Rules of logarithms
Each rule is an index law in disguise.
| Rule | Statement |
| Product | \(\log _a(MN)=\log _a M+\log _a N\) |
| Quotient | \(\log _a\!\left (\frac {M}{N}\right )=\log _a M-\log _a N\) |
| Power | \(\log _a\left (M^k\right )=k\log _a M\) |
| Special values | \(\log _a 1=0\), \(\log _a a=1\) |
| Change of base | \(\log _a x=\frac {\log _b x}{\log _b a}\) |
Note 6.9. There is no rule for \(\log (M+N)\). In particular \(\log (M+N)\) is not \(\log M+\log N\) — that expression equals \(\log (MN)\). Mixing these two up is the single commonest error in the chapter.
6.3.2 Logarithmic equations
Solution. (a) Rewrite in exponential form: \[\log _3 x=4\quad \implies \quad x=3^4=81.\]
(b) Combine the two logarithms first, using the product rule: \[\log _2\big [x(x-2)\big ]=3.\] Now convert to exponential form: \[x(x-2)=2^3=8\quad \implies \quad x^2-2x-8=0.\] Factorise: \[(x-4)(x+2)=0\quad \implies \quad x=4 \text { or } x=-2.\]
Now test both against the original equation. If \(x=-2\) then the equation contains \(\log _2(-2)\), which does not exist. So \(x=-2\) must be rejected, and \[x=4.\]
Check. \(\log _2 4+\log _2 2=2+1=3\). \(\relax \amscheckmark \)
Remark 6.11. That rejection is not a technicality — it is part of the answer, and marks are given for it. Combining logarithms can create an equation with solutions the original never had, because \(\log M+\log N\) requires \(M>0\) and \(N>0\) separately, while \(\log (MN)\) only requires their product to be positive. Two negatives would satisfy the second and not the first.
So the rule is: solve the combined equation, then substitute every root back into the original and discard any that ask for the logarithm of a non-positive number.
Solution. The base is \(10\), and the right-hand side is \(1=\log 10\). Combining the left side: \[\log \big [x(x+3)\big ]=1\quad \implies \quad x(x+3)=10^1=10\] \[\implies \quad x^2+3x-10=0\quad \implies \quad (x+5)(x-2)=0\] \[\implies \quad x=-5 \text { or } x=2.\]
Testing in the original: \(x=-5\) would require \(\log (-5)\), which does not exist, so it is rejected. Therefore \[x=2.\]
Check. \(\log 5+\log 2=\log 10=1\). \(\relax \amscheckmark \)
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.