Question

If $ {x^{2a - 3}}{y^{2a}} = {x^{6 - a}}{y^{5a}} $ then the value of a $ \log (\dfrac{x}{y}) $ is
 $ A) $ $ 3\log x $
 $ B) $ $ \log x $
 $ C) $ $ 6\log x $
 $ D) $ \[5\log x\]

Answer 1

Hint: First we have to define what the terms we need to solve the problem are.
These questions are based on the concept of the logarithm. A log function is defined as $ f(x) = {\log _b}x $ where log base is b since base e and \[10\] are commonly used and also the division term log is bottom to top of the multiplication.

Complete step by step answer:
Since the given question is of multiplication on \[x\] and \[y\] terms so we separate the x terms on left hand side and then separate the right-hand side in the right-hand side so that we simply apply the log formula and find the value;
Thus, $ {x^{2a - 3}}{y^{2a}} = {x^{6 - a}}{y^{5a}} \Rightarrow \dfrac{{{x^{2a - 3}}}}{{{x^{6 - a}}}} = \dfrac{{{y^{5a}}}}{{{y^{2a}}}} $ which is we separated the same terms on same side;
Now turn the denominator into numerator which yields using the property of division $ {x^{2a - 3}}{y^{2a}} = {x^{6 - a}}{y^{5a}} \Rightarrow \dfrac{{{x^{2a - 3}}}}{{{x^{6 - a}}}} = \dfrac{{{y^{5a}}}}{{{y^{2a}}}} \Rightarrow {x^{2a - 3 - (6 - a)}} = {y^{5a - 2a}} $ (turning the division terms as multiplication)
Now we going to apply the logarithm functions on both sides we get
 $ {x^{2a - 3 - (6 - a)}} = {y^{5a - 2a}} \Rightarrow 2a - 3 - (6 - a) \times \log x = (5a - 2a) \times \log y $ (On turning the log, the power part will become to the multiplication by the property of the log)
Hence further solving we get and also $ 2a - 3 - (6 - a) \times \log x = (5a - 2a) \times \log y \Rightarrow (3a - 9)\log x = (5a - 2a) \times \log y $
Thus, after simplifying we get $ \log \dfrac{x}{y} = \dfrac{3}{a}\log x $ since we are going to cross multiply the left- and right-hand side; $ \log \dfrac{x}{y} = \dfrac{3}{a}\log x \Rightarrow a\log \dfrac{x}{y} = 3\log x $ which is the required equation for the logarithm function.

So, the correct answer is “Option A”.

Note: $ \log {x^a} = a\log x $ and $ \dfrac{{{x^2}}}{{{x^1}}} = {x^{2 - 1}} = {x^1} $ (Turning the division terms as multiplication) these are some properties of logarithm that we used for this particular given problem, and for separate variable like \[x\] and \[y\] .