7. (i) Use logarithms to solve the equation \(8 ^ { 2 x + 1 } = 24\), giving your answer to 3 decimal places.
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Question 7:
Part (i):
Answer Marks
Guidance
Answer Mark
Guidance
\((2x+1)\log 8 = \log 24\) or \((2x+1) = \log_8 24\) M1
Takes logs and uses law of powers; any log base allowed
\(x = \frac{1}{2}\left(\frac{\log 24}{\log 8}-1\right)\) or \(x = \frac{1}{2}(\log_8 24 - 1)\) dM1
Makes \(x\) subject correctly
\(x = 0.264\) A1
Allow answers rounding to 0.264
Part (ii):
Answer Marks
Guidance
Answer Mark
Guidance
\(\log_2(11y-3) - \log_2 3 - \log_2 y^2 = 1\) M1
Applies power law replacing \(2\log_2 y\) with \(\log_2 y^2\)
\(\log_2\frac{(11y-3)}{3y^2} = 1\) dM1
Applies quotient/product law correctly to all three log terms
\(\log_2\frac{(11y-3)}{3y^2} = \log_2 2\) or \(\log_2\frac{(11y-3)}{y^2} = \log_2 6\) B1
States/uses \(\log_2 2 = 1\) or \(2^1 = 2\)
\(6y^2 - 11y + 3 = 0\) A1
Correct quadratic equation
Solves quadratic ddM1
Reasonable method
\(y = \frac{1}{3}\) and \(\frac{3}{2}\) (need both) A1
Neither should be rejected
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# Question 7:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(2x+1)\log 8 = \log 24$ or $(2x+1) = \log_8 24$ | M1 | Takes logs and uses law of powers; any log base allowed |
| $x = \frac{1}{2}\left(\frac{\log 24}{\log 8}-1\right)$ or $x = \frac{1}{2}(\log_8 24 - 1)$ | dM1 | Makes $x$ subject correctly |
| $x = 0.264$ | A1 | Allow answers rounding to 0.264 |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\log_2(11y-3) - \log_2 3 - \log_2 y^2 = 1$ | M1 | Applies power law replacing $2\log_2 y$ with $\log_2 y^2$ |
| $\log_2\frac{(11y-3)}{3y^2} = 1$ | dM1 | Applies quotient/product law correctly to all three log terms |
| $\log_2\frac{(11y-3)}{3y^2} = \log_2 2$ or $\log_2\frac{(11y-3)}{y^2} = \log_2 6$ | B1 | States/uses $\log_2 2 = 1$ or $2^1 = 2$ |
| $6y^2 - 11y + 3 = 0$ | A1 | Correct quadratic equation |
| Solves quadratic | ddM1 | Reasonable method |
| $y = \frac{1}{3}$ and $\frac{3}{2}$ (need both) | A1 | Neither should be rejected |
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7. (i) Use logarithms to solve the equation $8 ^ { 2 x + 1 } = 24$, giving your answer to 3 decimal places.\\
(ii) Find the values of $y$ such that
$$\log _ { 2 } ( 11 y - 3 ) - \log _ { 2 } 3 - 2 \log _ { 2 } y = 1 , \quad y > \frac { 3 } { 11 }$$
\hfill \mbox{\textit{Edexcel C2 2015 Q7 [9]}}