June 2023 Paper 3 Q1

Notes

M1*: Take logs of both sides (any base) correctly and use power law correctly at least once. Common correct answers that score M1 are: \(2x + 1 = \log_4 5^x\), \(\log_5 4^{2x + 1} = x\), \((2x + 1)\ln 4 = \ln(5^x)\)
Condone lack of bracket on the \(2x + 1\) term

M1dep*: Re-arrange to get an equation with a single term in \(x\) – condone sign errors only e.g. the following score the first two M marks: \(\dfrac{1}{x} + 2 = \dfrac{\log 5}{\log 4}\), \(x\left(2 - \dfrac{\log 5}{\log 4}\right) = -1\), \(x(\log_4 5 - 2) = 1\), \(x(1 - 2\log_5 4) = \log_5 4\), \(x(2\ln 4 - \ln 5) + \ln 4 = 0\)
This mark can be implied by a correct answer provided the first M mark was awarded (that is, we must see logs being taken and the power law used at least once)

A1: awrt \(-1.19\)
\(-1.19184404\ldots\)
Correct answer with no working scores no marks

Alternative method

Scheme	Marks
\(4^{2x + 1} = 5^x \Rightarrow \left(\dfrac{5}{16}\right)^x = 4\)	B1
\(x = \log_{\frac{5}{16}} 4\) or \(x\log\dfrac{5}{16} = \log 4\)	M1
\(x = -1.19\)	A1

B1: Correctly re-writes the given equation in the form \(a^x = b\) with \(a\) and \(b\) correct (check carefully for other equivalent correct equations)

M1: Taking logs correctly of their \(a^x = b\), where \(a\) and \(b\) are both positive, to obtain either \(x\log a = \log b\) (any base) or \(x = \log_a b\) (for their \(a\) and \(b\))
Not dependent on the B mark

A1: awrt \(-1.19\)
\(-1.19184404\ldots\)
Correct answer with no working scores no marks