TMUA: Exponential graphs and laws of logarithms
Exponential graphs and the laws of logarithms.
- Section 1 Part 1 · MM5 Exponentials and logarithms · spec MM5.1-MM5.2
- Papers 1 and 2
- 7 practice questions
- No calculator
What the specification covers
- y = a^x for positive a
- Laws of logs, log(1/x) = -log x, log_a a = 1
- No change-of-base questions are set (spec MM5.2)
Key ideas
- \(a^b=c\iff b=\log_ac\).
- \(\log_ax+\log_ay=\log_a(xy)\), \(\log_ax-\log_ay=\log_a\tfrac xy\), \(k\log_ax=\log_a(x^k)\).
- \(\log_a\tfrac1x=-\log_ax\) and \(\log_aa=1\).
- The TMUA specification says no questions need the change-of-base formula.
Common mistakes
- \(\log(x+y)\) does not split.
- \(\log_a(18a^2)\) includes \(2\log_aa=2\).
Exam tip
Break numbers into prime factors before applying the log laws.
Worked example
Worked example
Given that \(\log_a2=p\) and \(\log_a3=q\), which of the following is equal to \(\log_a(18a^2)\)?
- \(2p+2q\)
- \(a^2+p+2q\)
- \(2+p+2q\)
- \(2pq^2\)
- \(2+2p+q\)
Answer: C: \(2+p+2q\)
\(\log_a(18a^2)=\log_a2+2\log_a3+2\log_aa=p+2q+2\).
Practice questions
Try each question before opening the solution. No calculator: the TMUA does not allow one. Every answer here was re-checked independently by computer before it was published.
Question 1
Given that \(\log_a2=p\) and \(\log_a3=q\), which of the following is equal to \(\log_a(24a^{-1})\)?
- \(- 3 p q\)
- \(3 p + q - 1\)
- \(3 p + q + 1\)
- \(3 p + q\)
- \(p + 3 q - 1\)
Show the answer and solution
Answer: B: \(3 p + q - 1\)
\(24=2^{3}\times3^{1}\), so \(\log_a(24a^{-1})=3\log_a2+\log_a3-\log_aa=3 p + q - 1\).
Question 2
Given that \(\log_a2=p\) and \(\log_a3=q\), which of the following is equal to \(\log_a(24a^{3})\)?
- \(3 p + q\)
- \(p + 3 q + 3\)
- \(3 p + q + 3\)
- \(3 p + q - 3\)
- \(9 p q\)
Show the answer and solution
Answer: C: \(3 p + q + 3\)
\(24=2^{3}\times3^{1}\), so \(\log_a(24a^{3})=3\log_a2+\log_a3+3\log_aa=3 p + q + 3\).
Question 3
Given that \(\log_a2=p\) and \(\log_a3=q\), which of the following is equal to \(\log_a(18a^{2})\)?
- \(p + 2 q\)
- \(p + 2 q + 2\)
- \(2 p + q + 2\)
- \(4 p q\)
- \(p + 2 q - 2\)
Show the answer and solution
Answer: B: \(p + 2 q + 2\)
\(18=2^{1}\times3^{2}\), so \(\log_a(18a^{2})=\log_a2+2\log_a3+2\log_aa=p + 2 q + 2\).
Question 4
Given that \(\log_a2=p\) and \(\log_a3=q\), which of the following is equal to \(\log_a(12a^{-1})\)?
- \(- 2 p q\)
- \(p + 2 q - 1\)
- \(2 p + q + 1\)
- \(2 p + q - 1\)
- \(2 p + q\)
Show the answer and solution
Answer: D: \(2 p + q - 1\)
\(12=2^{2}\times3^{1}\), so \(\log_a(12a^{-1})=2\log_a2+\log_a3-\log_aa=2 p + q - 1\).
Question 5
Given that \(\log_a2=p\) and \(\log_a3=q\), which of the following is equal to \(\log_a(18a^{3})\)?
- \(p + 2 q + 3\)
- \(p + 2 q\)
- \(6 p q\)
- \(p + 2 q - 3\)
- \(2 p + q + 3\)
Show the answer and solution
Answer: A: \(p + 2 q + 3\)
\(18=2^{1}\times3^{2}\), so \(\log_a(18a^{3})=\log_a2+2\log_a3+3\log_aa=p + 2 q + 3\).
Question 6
Given that \(\log_a2=p\) and \(\log_a3=q\), which of the following is equal to \(\log_a(24a^{2})\)?
- \(3 p + q + 2\)
- \(3 p + q - 2\)
- \(p + 3 q + 2\)
- \(6 p q\)
- \(3 p + q\)
Show the answer and solution
Answer: A: \(3 p + q + 2\)
\(24=2^{3}\times3^{1}\), so \(\log_a(24a^{2})=3\log_a2+\log_a3+2\log_aa=3 p + q + 2\).
Keep going
- Previous topic: Trig graphs, identities and equations
- Next topic: Exponential equations
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