The answer
(a) shown
(b) \(x^3 + 2x^2 - 81 = 0\)
O-Level A-Math 2021 Paper 1 Question 12 · Verified worked solution by the Genius Plus Academy teaching team
The question
(a) Show the solution of \(6^x = 5\times 3^{x+1}\) is \(x = \dfrac{\lg 15}{\lg 2}\). [4]
(b) Express \(\log_3 x + \log_9(x + 2) = 2\) as a cubic equation in \(x\). [4]
(a) \(6^x = 5\times 3^{x+1} = 5\times 3\times 3^x = 15\times 3^x\). Dividing both sides by \(3^x\): \(\dfrac{6^x}{3^x} = 2^x = 15\). Taking \(\lg\): \(x\lg 2 = \lg 15 \Rightarrow x = \dfrac{\lg 15}{\lg 2}\). (shown)
(b) Change the base of the second log using \(\log_9(x + 2) = \dfrac{\log_3(x + 2)}{\log_3 9} = \tfrac12\log_3(x + 2)\). Then \[\log_3 x + \tfrac12\log_3(x + 2) = 2 \Rightarrow 2\log_3 x + \log_3(x + 2) = 4 \Rightarrow \log_3\!\big[x^2(x + 2)\big] = 4.\] So \(x^2(x + 2) = 3^4 = 81\), giving the cubic \(x^3 + 2x^2 - 81 = 0\).
Answer: (a) shown
(b) \(x^3 + 2x^2 - 81 = 0\)
Same structure, different numbers
Swap the constants, dress a quadratic as a length, hide a derivative inside an integral, and a student sees a brand new problem. The structure underneath is the same, and so is the method. Once a student can name the structure, a whole row of questions that look different start to open the same way.
That is where marks really leak: in choosing the method, not in the algebra that follows. We call it Lock and Key, name the lock, then the key follows.
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Genius Plus Academy · O-Level & IP Mathematics
Our O-Level A-Math tuition trains the same recognise-the-structure method these worked solutions show, taught by a team that has marked these papers for years. It runs within our weekly Secondary Math programme, Sec 1 to 4 and IP.
It is a index laws question from Exponential & logarithmic, worth 8 marks: 4 + 4.
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