Rearrange the first equation.

$\log_{2} (y + 1) + 2 \log_{2} x = 3$

Rewrite using logarithm laws.

$\log_{2} (y + 1) + \log_{2} x^{2} = 3$

Combine using logarithm laws.

$\log_{2} ((y + 1) (x^{2})) = 3$

Rewrite using logarithm laws.

$(y + 1) (x^{2}) = 2^{3} (y + 1) (x^{2}) = 8$

[1]

Rearrange the second equation.

$\log_{2} (x + 2) - \log_{2} y = 2$

Rewrite using logarithm laws.

$\log_{2} (\frac{x + 2}{y}) = 2$

Rewrite using logarithm laws.

$\begin{array}{rcl} \frac{x + 2}{y} & = & 2^{2} \\ x + 2 & = & 4 y \end{array}$

[1]

Solve the two equations $(y + 1) (x^{2}) = 8$ and $x + 2 = 4 y$ simultaneously. For example, by rearranging the second equation and using a substitution method.

$y = \frac{x + 2}{4}$

Substitute into the first equation.

$(x^{2}) (\frac{x + 2}{4} + 1) = 8$

[1]

Expand the brackets.

$\begin{array}{rcl} \frac{(x^{2}) (x + 2)}{4} + x^{2} & = & 8 \\ \frac{x^{3} + 2 x^{2}}{4} + x^{2} & = & 8 \end{array}$

Multiply through by 4.

$x^{3} + 2 x^{2} + 4 x^{2} = 32$

Simplify and rearrange to write in the form specified in the question.

$x^{3} + 6 x^{2} = 32$

$x^{3} + 6 x^{2} - 32 = 0$ [1]

Logarithmic & Exponential Functions (Cambridge O Level Additional Maths)

Exam Questions