Question

Given that \[{{\log }_{7}}{{x}^{2}}y=p\] and \[{{\log }_{7}}x{{y}^{2}}=q\], how do you find \[{{\log }_{7}}\sqrt[3]{xy}\] in terms of p and q?

Answer 1

Hint: This type of problem is based on the concept of logarithmic functions. We have to first add p and q. use the property of logarithm \[\log a+\log b=\log ab\]. Do necessary calculations in the obtained expression and simplify the equation using the property \[\log {{a}^{n}}=n\log a\]. Then, find the value of \[{{\log }_{7}}xy\]. Now, consider \[{{\log }_{7}}\sqrt[3]{xy}\] which can be written as \[{{\log }_{7}}{{\left( xy \right)}^{\dfrac{1}{3}}}\]. Using the property of logarithm, we get \[{{\log }_{7}}\sqrt[3]{xy}=\dfrac{1}{3}{{\log }_{7}}xy\]. Substitute the value of \[{{\log }_{7}}xy\] in terms of p and q obtained before. Do necessary calculations and find the value of \[{{\log }_{7}}\sqrt[3]{xy}\].

Complete step by step solution:
According to the question, we are asked to find the value of \[{{\log }_{7}}\sqrt[3]{xy}\] in terms of p and q.
We have been given \[{{\log }_{7}}{{x}^{2}}y=p\] ------------(1)
And \[{{\log }_{7}}x{{y}^{2}}=q\]. ----------(2)
Let us first add equation (1) and (2).
\[\Rightarrow {{\log }_{7}}{{x}^{2}}y+{{\log }_{7}}x{{y}^{2}}=p+q\]
We know that \[\log a+\log b=\log ab\]. Using this property of addition of logarithmic functions, we get
\[{{\log }_{7}}\left( {{x}^{2}}y\cdot x{{y}^{2}} \right)=p+q\]
On further simplification, we get
\[{{\log }_{7}}\left( {{x}^{3}}{{y}^{3}} \right)=p+q\]
On taking the cube common from x and y, we get
\[{{\log }_{7}}\left( {{\left( xy \right)}^{3}} \right)=p+q\]
We know that \[\log {{a}^{n}}=n\log a\]. Let us use this property of logarithm in the above expression.
\[\Rightarrow 3{{\log }_{7}}xy=p+q\]
Divide the whole equation by 3.
\[\Rightarrow \dfrac{3{{\log }_{7}}xy}{3}=\dfrac{p+q}{3}\]
We find that 3 are common in both the numerator and denominator of the LHS. On cancelling 3, we get
\[{{\log }_{7}}xy=\dfrac{p+q}{3}\] ------------(3)
Now, let us consider \[{{\log }_{7}}\sqrt[3]{xy}\].
We know that nth root of a term can be written as \[\sqrt[n]{a}={{a}^{\dfrac{1}{n}}}\].
Here, we can write the 3rd root as
\[{{\log }_{7}}\sqrt[3]{xy}={{\log }_{7}}{{\left( xy \right)}^{\dfrac{1}{3}}}\]
Using the properties of power in logarithmic functions \[\log {{a}^{n}}=n\log a\], we get
\[\Rightarrow {{\log }_{7}}\sqrt[3]{xy}=\dfrac{1}{3}{{\log }_{7}}xy\]
But from equation (3), we know that the value of \[{{\log }_{7}}xy\] is \[{{\log }_{7}}xy=\dfrac{p+q}{3}\].
On substituting in the above expression, we get
\[{{\log }_{7}}\sqrt[3]{xy}=\dfrac{1}{3}\left( \dfrac{p+q}{3} \right)\]
On further simplification, we get
\[{{\log }_{7}}\sqrt[3]{xy}=\left( \dfrac{p+q}{9} \right)\]
\[\therefore {{\log }_{7}}\sqrt[3]{xy}=\dfrac{p+q}{9}\]

Hence, we value of \[{{\log }_{7}}\sqrt[3]{xy}\] on terms of p and q is \[\dfrac{p+q}{9}\].

Note: We should be very thorough with the properties of logarithmic functions to solve this type of question. We can add two logarithmic functions and use these properties only when the bases are the same. Here the base was 7 and thus we could add the functions. Avoid calculation mistakes based on sign conventions.