Question

How to express in terms of $p$ and $q?$ Given ${\log _7}{x^2}y = p$ and ${\log _7}{x^2}{y^4} = 2q$, express ${\log _7}x{y^{\dfrac{1}{2}}}$ in terms of $p$ and $q$ .

Answer 1

Hint : In the given question we have to convert ${\log _7}x{y^{\dfrac{1}{2}}}$ in terms of $p$ and $q$ where ${\log _7}{x^2}y = p$ and ${\log _7}{x^2}{y^4} = 2q$ . This means that the final answer should only contain $p$ and $q$ . Here try to make the required term using both the given value of $p$ and $q$ by using basic operations of mathematics. Use different identities of logarithm functions to get the given term. You should recall all the logarithm function identities because you need many different identities to solve this question.

Complete step by step solution:
In the given question we have to express ${\log _7}x{y^{\dfrac{1}{2}}}$ in terms of $p$ and $q$ where,

$${\log _7}{x^2}y = p........(1) \\\ {\log _7}{x^2}{y^4} = 2q........(2) \;$$

Now we will try to get the given term ${\log _7}x{y^{\dfrac{1}{2}}}$ on the left hand side by using the value of $p$ and $q$ .
Now subtracting $(2)$ from $(1)$ we get,
${\log _7}{x^2}{y^4} - {\log _7}{x^2}y = (2q - p)$
Using logarithm identity $\left( {\log a - \log b = \log \dfrac{a}{b}} \right)$ we get:

$${\log _7}\dfrac{{{x^2}{y^4}}}{{{x^2}y}} = 2q - p \\\ \Rightarrow {\log _7}{y^3} = 2q - p \;$$

Now using logarithm identity $(\log {a^b} = b\log a)$ we get,

$$3{\log _7}y = 2q - p \\\ \Rightarrow {\log _7}y = \dfrac{{2q - q}}{3}................(3) \;$$

Multiplying both sides with $\dfrac{1}{2}$ we get:
$\dfrac{1}{2}{\log _7}y = \dfrac{{2q - p}}{6}$
Now using identity $(a\log y = \log {y^a})$ we get:
${\log _7}{y^{\dfrac{1}{2}}} = \dfrac{{2q - p}}{6}.............(4)$
Now equation $(1)$ can be written as:

$${\log _7}{x^2} + {\log _7}y = p \\\ \Rightarrow {\log _7}{x^2} = p - {\log _7}y \;$$

Now putting value of ${\log _7}y$ from equation $(3)$ we get:
${\log _7}{x^2} = p - \dfrac{{2q - p}}{3} = \dfrac{{4p - 2q}}{3}$
Using logarithm identity $(\log {y^a} = a\log y)$ we get:

$$2{\log _7}x = \dfrac{{4p - 2q}}{3} \\\ \Rightarrow {\log _7}x = \dfrac{{4p - 2q}}{3} \times \dfrac{1}{2} = \dfrac{{2q - p}}{3}...........(5) \;$$

Now adding the equations $(5)$ and $(4)$ we get,
${\log _7}x + \log {y^{\dfrac{1}{2}}} = \dfrac{{2q - p}}{3} + \dfrac{{2q - p}}{6}$
Using identity $(\log a + \log b = \log ab)$ we get:
${\log _7}x{y^{\dfrac{1}{2}}} = \dfrac{{6q - 3p}}{6} = \dfrac{{2q - p}}{2}$
Hence, ${\log _7}x{y^{\dfrac{1}{2}}}$ can be expressed as $\dfrac{{2q - p}}{2}$ .
So, the correct answer is “ $\dfrac{{2q - p}}{2}$”.

Note : Here you have to be very careful with the power as well base of the logarithm function. Also these questions can be solved with different methods but the above mentioned is the best and easiest method to solve these kinds of questions.
In these kinds of questions we have used many logarithm identities so you should be aware of all the logarithm identities basic as well as advanced. Here I am mentioning different logarithm identities which you should learn:

$$\log ab = \log a + \log b \\\ \log \dfrac{a}{b} = \log a - \log b \\\ \log {a^b} = b\log a \\\ {\log _b}a = \dfrac{{{{\log }_d}a}}{{{{\log }_d}b}} \;$$

These are only basic identities there are many more identities so learn all those to solve these kinds of questions easily.