0606 P23 - Jun 2021 - Q3 - 7 marks
7999
(a) Write
\(\frac{x(27xy^3)^{\frac53}}{\sqrt[4]{81y^5}}\)
in the form \(3^a x^b y^c\), where \(a\), \(b\) and \(c\) are constants.
(b)
(i) Find the value of \(a\) such that
\(2\log_a8=\frac32.\)
(ii) Write \(\log_{a^2}3a\) as a single logarithm to base \(a\).
Solution
Answer: (a) \(3^4x^{\frac83}y^{\frac{15}{4}}\); (b)(i) \(a=16\); (b)(ii) \(\log_a\sqrt{3a}\).
(a) First simplify the numerator:
\(x(27xy^3)^{\frac53} =x\cdot27^{\frac53}x^{\frac53}y^5.\)
Since \(27=3^3\),
\(27^{\frac53}=3^5.\)
So the numerator is
\(3^5x^{\frac83}y^5.\)
The denominator is
\(\sqrt[4]{81y^5}=3y^{\frac54}.\)
Therefore
\(\frac{x(27xy^3)^{\frac53}}{\sqrt[4]{81y^5}} =3^4x^{\frac83}y^{5-\frac54} =3^4x^{\frac83}y^{\frac{15}{4}}.\)
(b)(i) From
\(2\log_a8=\frac32,\)
we get
\(\log_a8=\frac34.\)
Therefore
\(a^{\frac34}=8.\)
Raise both sides to the power \(\frac43\):
\(a=8^{\frac43}=16.\)
(b)(ii) Use change of base:
\(\log_{a^2}3a=\frac{\log_a(3a)}{\log_a(a^2)}.\)
Since \(\log_a(a^2)=2\),
\(\log_{a^2}3a=\frac12\log_a(3a).\)
So
\(\log_{a^2}3a=\log_a(3a)^{\frac12} =\log_a\sqrt{3a}.\)
