WJEC Unit 1 Specimen — Question 12 3 marks

Exam BoardWJEC
ModuleUnit 1 (Unit 1)
SessionSpecimen
Marks3
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TopicLaws of Logarithms
TypeSimplify or prove logarithmic identity
DifficultyEasy -1.8 This is a trivial logarithm identity question requiring only the basic fact that log_a(a) = 1, making the left side simply 1 × log_a(19) = log_a(19). It's a one-step proof with no problem-solving required, significantly easier than typical A-level questions.
Spec1.01a Proof: structure of mathematical proof and logical steps1.06f Laws of logarithms: addition, subtraction, power rules
Prove that $$\log_a a \times \log_a 19 = \log_a 19$$ whatever the value of the positive constant \(a\). [3]
AnswerMarks Guidance
Let \(p = \log_a 19\), \(q = \log_a a\)B1 AO2 (the relationship between log and power)
Then \(19 = a^p\), \(a = 7^q\)B1 AO2 (the laws of indices)
\(19 = a^p = (7^q)^p = 7^{qp}\)— —
\(qp = \log_7 19\)— — (the relationship between log and power)
\(\log_7 a \times \log_a 19 = \log_7 19\)B1 AO2 (convincing)
Total: [3]
Let $p = \log_a 19$, $q = \log_a a$ | B1 | AO2 (the relationship between log and power)
Then $19 = a^p$, $a = 7^q$ | B1 | AO2 (the laws of indices)

$19 = a^p = (7^q)^p = 7^{qp}$ | — | — 

$qp = \log_7 19$ | — | — (the relationship between log and power)

$\log_7 a \times \log_a 19 = \log_7 19$ | B1 | AO2 (convincing)

**Total: [3]**

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Prove that
$$\log_a a \times \log_a 19 = \log_a 19$$
whatever the value of the positive constant $a$. [3]

\hfill \mbox{\textit{WJEC Unit 1  Q12 [3]}}
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Q1 7 Q2 6 Q3 6 Q4 5 Q5 12 Q6 5 Q7 5 Q8 6 Q9 7 Q10 8 Q11 3 Q12 3 Q13 7 Q14 8 Q15 8 Q16 5 Q17 12 Q18 7