8. (a) Given that
$$\log _ { 2 } ( y - 1 ) = 1 + \log _ { 2 } x ,$$
show that
$$y = 2 x + 1 .$$
(b) Solve the simultaneous equations
$$\begin{aligned}
& \log _ { 2 } ( y - 1 ) = 1 + \log _ { 2 } x \\
& 2 \log _ { 3 } y = 2 + \log _ { 3 } x
\end{aligned}$$
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Part (a)
Answer Marks
\(\log_2 (y - 1) - \log_2 x = 1, \quad \log_2 \frac{y-1}{x} = 1\) M1
\(\frac{y-1}{x} = 2^1 = 2\) M1
\(y - 1 = 2x, \quad y = 2x + 1\) A1
Part (b)
Answer Marks
\(2 \log_3 y = 2 + \log_3 x \Rightarrow \log_3 y^2 - \log_3 x = 2\) M1
\(\frac{y^2}{x} = 3^2 = 9\) M1
\(y^2 = 9x\) A1
sub. \(y = 2x + 1\): \((2x + 1)^2 = 9x\) M1
\(4x^2 + 4x + 1 = 9x\) \(4x^2 - 5x + 1 = 0\) \((4x - 1)(x - 1) = 0\) M1
\(x = \frac{1}{4}, 1\) A1
\(\therefore x = \frac{1}{4}, y = \frac{3}{2}\) or \(x = 1, y = 3\) A1
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**Part (a)**
$\log_2 (y - 1) - \log_2 x = 1, \quad \log_2 \frac{y-1}{x} = 1$ | M1 |
$\frac{y-1}{x} = 2^1 = 2$ | M1 |
$y - 1 = 2x, \quad y = 2x + 1$ | A1 |
**Part (b)**
$2 \log_3 y = 2 + \log_3 x \Rightarrow \log_3 y^2 - \log_3 x = 2$ | M1 |
$\frac{y^2}{x} = 3^2 = 9$ | M1 |
$y^2 = 9x$ | A1 |
sub. $y = 2x + 1$: $(2x + 1)^2 = 9x$ | M1 |
$4x^2 + 4x + 1 = 9x$ |
$4x^2 - 5x + 1 = 0$ |
$(4x - 1)(x - 1) = 0$ | M1 |
$x = \frac{1}{4}, 1$ | A1 |
$\therefore x = \frac{1}{4}, y = \frac{3}{2}$ or $x = 1, y = 3$ | A1 |
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8. (a) Given that
$$\log _ { 2 } ( y - 1 ) = 1 + \log _ { 2 } x ,$$
show that
$$y = 2 x + 1 .$$
(b) Solve the simultaneous equations
$$\begin{aligned}
& \log _ { 2 } ( y - 1 ) = 1 + \log _ { 2 } x \\
& 2 \log _ { 3 } y = 2 + \log _ { 3 } x
\end{aligned}$$
\hfill \mbox{\textit{Edexcel C2 Q8 [10]}}