Moderate -0.3 This is a straightforward modulus equation requiring students to split into two cases (2^x - 7 = 1 or 2^x - 7 = -1), then solve simple exponential equations using logarithms. While it requires understanding of modulus and exponentials, it's a standard textbook exercise with clear methodology and minimal steps, making it slightly easier than average.
State or imply non-modular equation \((2^x - 7)^2 = 1^2\), or corresponding pair of equations
M1
Obtain \(2^x = 8\) and \(2^x = 6\)
A1
State answer 3
B1
Use logarithmic method to solve an equation of the form \(2^x = k\), where \(k > 0\)
M1
State answer 2.58
A1
[5]
Or branch:
Answer
Marks
Guidance
State or imply one value for \(2^x\), e.g. 8, by solving an equation or by inspection
B1
State answer 3
B1
State second value for \(2^x\)
B1
Use logarithmic method to solve an equation of the form \(2^x = k\), where \(k > 0\)
M1
State answer 2.58
A1
[5]
Either branch:
State or imply non-modular equation $(2^x - 7)^2 = 1^2$, or corresponding pair of equations | M1 |
Obtain $2^x = 8$ and $2^x = 6$ | A1 |
State answer 3 | B1 |
Use logarithmic method to solve an equation of the form $2^x = k$, where $k > 0$ | M1 |
State answer 2.58 | A1 | [5]
Or branch:
State or imply one value for $2^x$, e.g. 8, by solving an equation or by inspection | B1 |
State answer 3 | B1 |
State second value for $2^x$ | B1 |
Use logarithmic method to solve an equation of the form $2^x = k$, where $k > 0$ | M1 |
State answer 2.58 | A1 | [5]