Question 13:
(a) Express y in terms of x and / or e of
(i) 6 ln y = √5 ln x,
(ii) ln (x + y2 ) – 3 = 0.
[ 4 marks]
(b) Diagram 9 shows the length of the chord BC at semicircle ABCD.
It is given that the length of AD is (x + 2√2)cm and the ratio of the radius of the semicircle to the length of the chord is 6 : 1.
Find the value of x in the form a + b√2, such that a and b are rational numbers. [ 4 marks]
Solution:
(a)(i)
$$ \begin{aligned} 6 \ln y & =\sqrt{5} \ln x \\ 6 \log _e y & =\sqrt{5} \log _e x \\ \log _e y^6 & =\log _e x^{\sqrt{5}} \\ y^6 & =x^{\sqrt{5}} \\ y & =\sqrt[6]{x^{\sqrt{5}}} \\ y & =\sqrt[12]{x^{\sqrt{5}}} \end{aligned} $$
(a)(ii)
$$ \begin{aligned} \ln \left(x+y^2\right)-3 & =0 \\ \ln \left(x+y^2\right) & =3 \\ \log _e\left(x+y^2\right) & =3 \\ x+y^2 & =e^3 \\ y^2 & =e^3-x \\ y & =\sqrt{e^3-x} \end{aligned} $$
(b)
Given that diameter = x + 2√2
and ratio of radius : chord = 6 : 1
$$ \begin{aligned} & \frac{\text { Radius }}{\text { Chord }}=\frac{6}{1} \\ & \text { Radius }=6(\text { Chord }) \end{aligned} $$
$$ \begin{aligned} \frac{x+2 \sqrt{2}}{2} & =6(x \sqrt{2}-1) \\ x+2 \sqrt{2} & =12(x \sqrt{2}-1) \\ x+2 \sqrt{2} & =12 \sqrt{2} x-12 \\ 12 \sqrt{2} x-x & =2 \sqrt{2}+12 \\ x(12 \sqrt{2}-1) & =2 \sqrt{2}+12 \\ x & =\frac{2 \sqrt{2}+12}{12 \sqrt{2}-1} \end{aligned} $$
$$ \begin{aligned} &x=\frac{(2 \sqrt{2}+12)(12 \sqrt{2}+1)}{(12 \sqrt{2}-1)(12 \sqrt{2}+1)}\\ &x=\frac{24(2)+2 \sqrt{2}+144 \sqrt{2}+12}{144(2)-1}\\ &x=\frac{60+146 \sqrt{2}}{287}\\ &x=\frac{60}{287}+\frac{146 \sqrt{2}}{287} \end{aligned} $$
(a) Express y in terms of x and / or e of
(i) 6 ln y = √5 ln x,
(ii) ln (x + y2 ) – 3 = 0.
[ 4 marks]
(b) Diagram 9 shows the length of the chord BC at semicircle ABCD.
It is given that the length of AD is (x + 2√2)cm and the ratio of the radius of the semicircle to the length of the chord is 6 : 1.Find the value of x in the form a + b√2, such that a and b are rational numbers. [ 4 marks]
Solution:
(a)(i)
$$ \begin{aligned} 6 \ln y & =\sqrt{5} \ln x \\ 6 \log _e y & =\sqrt{5} \log _e x \\ \log _e y^6 & =\log _e x^{\sqrt{5}} \\ y^6 & =x^{\sqrt{5}} \\ y & =\sqrt[6]{x^{\sqrt{5}}} \\ y & =\sqrt[12]{x^{\sqrt{5}}} \end{aligned} $$
(a)(ii)
$$ \begin{aligned} \ln \left(x+y^2\right)-3 & =0 \\ \ln \left(x+y^2\right) & =3 \\ \log _e\left(x+y^2\right) & =3 \\ x+y^2 & =e^3 \\ y^2 & =e^3-x \\ y & =\sqrt{e^3-x} \end{aligned} $$
(b)
Given that diameter = x + 2√2
and ratio of radius : chord = 6 : 1
$$ \begin{aligned} & \frac{\text { Radius }}{\text { Chord }}=\frac{6}{1} \\ & \text { Radius }=6(\text { Chord }) \end{aligned} $$
$$ \begin{aligned} \frac{x+2 \sqrt{2}}{2} & =6(x \sqrt{2}-1) \\ x+2 \sqrt{2} & =12(x \sqrt{2}-1) \\ x+2 \sqrt{2} & =12 \sqrt{2} x-12 \\ 12 \sqrt{2} x-x & =2 \sqrt{2}+12 \\ x(12 \sqrt{2}-1) & =2 \sqrt{2}+12 \\ x & =\frac{2 \sqrt{2}+12}{12 \sqrt{2}-1} \end{aligned} $$
$$ \begin{aligned} &x=\frac{(2 \sqrt{2}+12)(12 \sqrt{2}+1)}{(12 \sqrt{2}-1)(12 \sqrt{2}+1)}\\ &x=\frac{24(2)+2 \sqrt{2}+144 \sqrt{2}+12}{144(2)-1}\\ &x=\frac{60+146 \sqrt{2}}{287}\\ &x=\frac{60}{287}+\frac{146 \sqrt{2}}{287} \end{aligned} $$