Easy -1.3 This is a straightforward index law question requiring only the recognition that 8 = 2³ and application of the power rule (aᵐ)ⁿ = aᵐⁿ. It's a single-step manipulation with no problem-solving required, making it easier than average but not completely trivial.
\((8^{2x+3}) = (2^3)^{2x+3} = 2^{3(2x+3)}\) or \(2^{ax+b}\) with \(a=6\) or \(b=9\)
M1
Uses \(8=2^3\) and multiplies powers \(3(2x+3)\). Does not add powers. Just \(8=2^3\) or \(8^{\frac{1}{3}}=2\) is M0.
\(= 2^{6x+9}\) or \(2^{3(2x+3)}\) as final answer with no errors, or \((y=)6x+9\) or \(3(2x+3)\)
A1
Either form accepted.
Alternative using logs: \(8^{2x+3}=2^y \Rightarrow (2x+3)\log 8 = y\log 2 \Rightarrow y=\dfrac{(2x+3)\log 8}{\log 2}\)
M1
So \((y=)6x+9\) or \(3(2x+3)\)
A1
Total: 2 marks
## Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(8^{2x+3}) = (2^3)^{2x+3} = 2^{3(2x+3)}$ or $2^{ax+b}$ with $a=6$ or $b=9$ | M1 | Uses $8=2^3$ and **multiplies** powers $3(2x+3)$. Does **not** add powers. Just $8=2^3$ or $8^{\frac{1}{3}}=2$ is M0. |
| $= 2^{6x+9}$ or $2^{3(2x+3)}$ as final answer with no errors, or $(y=)6x+9$ or $3(2x+3)$ | A1 | Either form accepted. |
**Alternative using logs:** $8^{2x+3}=2^y \Rightarrow (2x+3)\log 8 = y\log 2 \Rightarrow y=\dfrac{(2x+3)\log 8}{\log 2}$ | M1 | |
So $(y=)6x+9$ or $3(2x+3)$ | A1 | |
**Total: 2 marks**