Moderate -0.3 This is a straightforward application of logarithm laws (power rule, quotient rule, and converting log equations to exponential form). Both parts follow standard textbook patterns with no conceptual challenges—slightly easier than average due to the routine algebraic manipulation required, though the two-part structure and simplification steps keep it close to typical difficulty.
7. (i) \(2 \log ( x + a ) = \log \left( 16 a ^ { 6 } \right)\), where \(a\) is a positive constant Find \(x\) in terms of \(a\), giving your answer in its simplest form.
(ii) \(\quad \log _ { 3 } ( 9 y + b ) - \log _ { 3 } ( 2 y - b ) = 2\), where \(b\) is a positive constant Find \(y\) in terms of \(b\), giving your answer in its simplest form.
(i) Use of power rule so \(\log(x+a)^{2}=\log 16a^{6}\) or \(2\log(x+a)=2\log 4a^{3}\) or \(\log(x+a)=\log(16a^{6})^{\frac{1}{2}}\)
M1
Removes logs and square roots, or halves then removes logs to give \((x+a)=4a^{3}\)
M1
\((x=)4a^{3}-a\) (depends on previous M's and must be this expression or equivalent)
A1cao
(ii) Way 1: \(\log_{3}\frac{(9y+b)}{(2y-b)}=2\)
M1
\(\frac{(9y+b)}{(2y-b)}=3^{2}\)
M1
\((9y+b)=9(2y-b)\Rightarrow y=\)
M1
\(y=\frac{10}{9}b\)
A1cso
Way 2: Or: \(\log_{3}(9y+b)=\log_{3}9+\log_{3}(2y-b)\)
2nd M mark M1
\(\log_{3}(9y+b)=\log_{3}9(2y-b)\)
1st M mark M1
\((9y+b)=9(2y-b)\Rightarrow y=\)
M1
\(y=\frac{10}{9}b\)
A1cso
Notes:
- (i)1st M1: Applies power law of logarithms correctly to one side of the equation; M1: Correct log work in correct order. If they square and obtain a quadratic the algebra should be correct. The marks is for \(x+a=\sqrt{16a^{6}}\) isw so allow \(x+a=\pm 4a^{3}\) for Method mark. Also allow \(x+a=4a^{3}\) or \(x+a=\pm 4a^{2\frac{5}{6}}\) or even \(x+a=16a^{3}\) as there is evidence of attempted square root. May see correct \(x+a=10^{(\log 4+3\log a)}\) so \(x=-a+10^{(\log 4+3\log a)}\) which gains M1A0 unless followed by the answer in the scheme; A1: Do not allow \(x=\pm 4a^{3}-a\) for accuracy mark. You may see the factorised \(a(2a+1)(2a-1)\) o.e.
- (ii)M1: Applying the subtraction or addition law of logarithms correctly to make two log terms into one log term in \(y\); M1: Uses \(\log_{3}3^{2}=2\); 3rd M1: Obtains correct linear equation in \(y\) usually the one in the scheme and attempts \(y=\); A1cso: \(y=\frac{10}{9}b\) or correct equivalent after completely correct work.
- Special case:
\[\log_{3}\frac{(9y+b)}{(2y-b)}=2 \text{ is M0 unless clearly crossed out and replaced by the correct } \log_{3}\frac{(9y+b)}{(2y-b)}=2\]
Candidates may then write \(\frac{(9y+b)}{(2y-b)}=3^{2}\) and proceed to the correct answer – allow M0M1M1A0 as the answer requires a completely correct solution.
| (i) Use of power rule so $\log(x+a)^{2}=\log 16a^{6}$ or $2\log(x+a)=2\log 4a^{3}$ or $\log(x+a)=\log(16a^{6})^{\frac{1}{2}}$ | M1 |
|---|---|
| Removes logs and square roots, or halves then removes logs to give $(x+a)=4a^{3}$ | M1 |
| $(x=)4a^{3}-a$ (depends on previous M's and must be this expression or equivalent) | A1cao |
| (ii) **Way 1:** $\log_{3}\frac{(9y+b)}{(2y-b)}=2$ | M1 |
| $\frac{(9y+b)}{(2y-b)}=3^{2}$ | M1 |
| $(9y+b)=9(2y-b)\Rightarrow y=$ | M1 |
| $y=\frac{10}{9}b$ | A1cso |
| **Way 2:** Or: $\log_{3}(9y+b)=\log_{3}9+\log_{3}(2y-b)$ | 2nd M mark M1 |
| $\log_{3}(9y+b)=\log_{3}9(2y-b)$ | 1st M mark M1 |
| $(9y+b)=9(2y-b)\Rightarrow y=$ | M1 |
| $y=\frac{10}{9}b$ | A1cso |
**Notes:**
- **(i)** **1st M1:** Applies power law of logarithms correctly to one side of the equation; M1: Correct log work in correct order. If they square and obtain a quadratic the algebra should be correct. The marks is for $x+a=\sqrt{16a^{6}}$ isw so allow $x+a=\pm 4a^{3}$ for Method mark. Also allow $x+a=4a^{3}$ or $x+a=\pm 4a^{2\frac{5}{6}}$ or even $x+a=16a^{3}$ as there is evidence of attempted square root. May see correct $x+a=10^{(\log 4+3\log a)}$ so $x=-a+10^{(\log 4+3\log a)}$ which gains M1A0 unless followed by the answer in the scheme; **A1:** Do not allow $x=\pm 4a^{3}-a$ for accuracy mark. You may see the factorised $a(2a+1)(2a-1)$ o.e.
- **(ii)** **M1:** Applying the subtraction or addition law of logarithms correctly to make **two log terms into one log term in** $y$; M1: Uses $\log_{3}3^{2}=2$; **3rd M1:** Obtains **correct linear equation in** $y$ usually the one in the scheme and attempts $y=$; **A1cso:** $y=\frac{10}{9}b$ or correct equivalent after **completely correct work**.
- **Special case:**
$$\log_{3}\frac{(9y+b)}{(2y-b)}=2 \text{ is M0 unless clearly crossed out and replaced by the correct } \log_{3}\frac{(9y+b)}{(2y-b)}=2$$
Candidates may then write $\frac{(9y+b)}{(2y-b)}=3^{2}$ and proceed to the **correct answer** – allow M0M1M1A0 as the answer requires a completely correct solution.
7. (i) $2 \log ( x + a ) = \log \left( 16 a ^ { 6 } \right)$, where $a$ is a positive constant Find $x$ in terms of $a$, giving your answer in its simplest form.\\
(ii) $\quad \log _ { 3 } ( 9 y + b ) - \log _ { 3 } ( 2 y - b ) = 2$, where $b$ is a positive constant Find $y$ in terms of $b$, giving your answer in its simplest form.\\
\hfill \mbox{\textit{Edexcel C2 2017 Q7 [7]}}