# 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

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**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 \[y\] .

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