Short Questions (Question 15 – 17)

Posted on May 18, 2020 by Myhometuition

Question 15 (4 marks):

$\begin{array}{l} (a) Given P = \log_{a} Q, state the \\ conditions of a . \\ (b) Given \log_{3} y = \frac{2}{\log_{x y} 3}, express \\ y in terms of x . \end{array}$

Solution:
(a)
a > 0, a ≠ 1

(b)

$\begin{array}{l} \log_{3} y = \frac{2}{\log_{x y} 3} \\ \frac{\log_{x y} y}{\log_{x y} 3} = \frac{2}{\log_{x y} 3} \\ \log_{x y} y = 2 \\ y = {(x y)}^{2} \\ y = x^{2} y^{2} \\ \frac{1}{x^{2}} = \frac{y^{2}}{y} \\ y = \frac{1}{x^{2}} \end{array}$

Question 16 (3 marks):

$Given \frac{25^{h + 3}}{125^{p - 1}} =1, express p in terms of h .$

Solution:

$\begin{array}{l} \frac{25^{h + 3}}{125^{p - 1}} = 1 \\ 25^{h + 3} = 125^{p - 1} \\ {(5^{2})}^{h + 3} = {(5^{3})}^{p - 1} \\ 5^{2 h + 6} = 5^{3 p - 3} \\ 2 h + 6 = 3 p - 3 \\ 3 p = 2 h + 9 \\ p = \frac{2 h + 9}{3} \end{array}$

Question 17 (3 marks):

$\begin{array}{l} Solve the equation: \\ \log_{m} 324 - \log_{\sqrt{m}} 2 m = 2 \end{array}$

Solution:

$\begin{array}{l} \log_{m} 324 - \log_{\sqrt{m}} 2 m = 2 \\ \log_{m} 324 - \frac{\log_{m} 2 m}{\log_{m} m^{\frac{1}{2}}} = 2 \\ \log_{m} 324 - 2 (\frac{\log_{m} 2 m}{\log_{m} m}) = 2 \\ \log_{m} 324 - 2 \log_{m} 2 m = 2 \\ \log_{m} 324 - \log_{m} {(2 m)}^{2} = l o g_{m} m^{2} \\ \log_{m} (\frac{324}{4 m^{2}}) = l o g_{m} m^{2} \\ \frac{324}{4 m^{2}} = m^{2} \\ 4 m^{4} = 324 \\ m^{4} = 81 \\ m = \pm 3 (- 3 is rejected) \end{array}$