OMPT drill — Functions

Logarithms

Log rules, change of base, and solving logarithmic equations. In my sessions this is the topic with the widest gap between "seen it" and "can do it under time". Attempt each problem before opening the solution; the notation only becomes automatic by writing it yourself.

Tested inOMPT-A OMPT-B OMPT-D OMPT-E OMPT-F OMPT-G

Lesson

A logarithm answers one question: to what power must I raise the base to get this number? $lo g_{2} 32$ asks "2 to what power gives 32?" and the answer is 5. That is the entire definition, and I encourage students to say the sentence out loud every time, because the notation intimidates people into forgetting how little it means. $lo g_{b} a = c$ and $b^{c} = a$ are the same statement wearing different clothes.

In my experience this is the topic with the widest gap between recognising and doing. Everyone has seen the log rules; far fewer can deploy them at speed, in the right direction, under a clock. The rules themselves are just the power rules from the exponents chapter read backwards: logs turn multiplication into addition because exponents turn addition into multiplication. Hold onto that connection and the rules stop being three arbitrary facts to memorise.

Problem 1Numeric answer

Evaluate $lo g_{3} 81$ .

Show the worked solution

Answer: 4

The logarithm asks: 3 to which power gives 81? Since $3^{4} = 81$ , the answer is 4. When I blank on a log, I rewrite it as its exponential twin: $lo g_{3} 81 = x$ means $3^{x} = 81$ .

Problem 2Multiple choice

For positive numbers $a$ and $b$ , which expression equals $lo g a + lo g b$ ?

A $lo g (ab)$
B $lo g (a + b)$
C $lo g a \cdot lo g b$
D $lo g (a^{b})$

Show the worked solution

Answer: A — $lo g (ab)$

Adding logs multiplies their arguments: $lo g a + lo g b = lo g (ab)$ . This mirrors the exponent rule $1 0^{x} \cdot 1 0^{y} = 1 0^{x + y}$ — logs turn multiplication into addition, which is the whole reason they were invented.

Problem 3Numeric answer

Solve for $x$ : $lo g_{2} (x - 1) = 5$ .

Show the worked solution

Answer: 33

I convert to exponential form: $x - 1 = 2^{5} = 32$ , so $x = 33$ . The argument is positive ( $33 - 1 = 32 > 0$ ), so the solution is valid. That domain check takes two seconds and catches phantom solutions in harder problems.

Problem 4Multiple choice

Given $lo g 2 \approx 0.301$ , what is $lo g 8$ approximately?

A $0.903$
B $0.602$
C $1.204$
D $0.027$

Show the worked solution

Answer: A — $0.903$

Since $8 = 2^{3}$ , the power rule gives $lo g 8 = 3 lo g 2 \approx 3 \cdot 0.301 = 0.903$ . No calculator needed once you spot that 8 is a power of 2.

Problem 5Spot the error

A student simplifies. Step 1: $lo g (x + y) = lo g x + lo g y$ . Step 2: with $x = 10$ , $y = 90$ , the left side becomes $lo g 100 = 2$ . Step 3: the right side becomes $1 + lo g 90 \approx 2.95$ . Which step contains the error?

Show the worked solution

Answer: Step 1

Step 1 states a rule that does not exist: the log of a sum does not split. Steps 2 and 3 actually expose the problem, since the two sides disagree ( $2 \neq = 2.95$ ). The genuine rule concerns products: $lo g (x y) = lo g x + lo g y$ , and indeed $lo g (10 \cdot 90) = lo g 900 \approx 2.95$ matches the right side.

Problem 6Numeric answer

Solve for $k$ : $1 0^{2 k} = 1000$ .

Show the worked solution

Answer: 1.5

Taking $lo g_{10}$ of both sides — or simply writing $1000 = 1 0^{3}$ — gives $2 k = 3$ , so $k = 1.5$ . Check: $1 0^{3} = 1000$ .