Question

If $${\log }_{3}5=x$$and $${\log }_{25}11=y$$, then the value of $${\log }_3\left({\displaystyle \frac{11}{3}}\right)$$ in terms of $$x$$ and $$y$$ is

Answer 1

Hint: Here, we will use the logarithm properties like, $${\log }_{b^2}a={\displaystyle \frac{1}{2}}{\log }_{b}a$$ , $${\log }_{b}a={\displaystyle \frac{{\log }_{c}a}{{\log }_{c}b}}$$ and $${\log }_c\left({\displaystyle \frac{a}{b}}\right)={\log }_{c}a-{\log }_{c}b$$ to rewrite the given conditions in order to find the required value.

Complete step-by-step answer:
We are given that the $${\log }_{3}5=x$$ and $${\log }_{25}11=y$$.
We will now rewrite the expression $${\log }_{25}11=y$$, we get
$$\Rightarrow {\log }_{5^2}11=y$$
Using the logarithm property, $${\log }_{b^2}a={\displaystyle \frac{1}{2}}{\log }_{b}a$$ in the above expression, we get
$$\Rightarrow {\displaystyle \frac{1}{2}}{\log }_{5}11=y$$
Multiplying the above equation by 2 on both sides, we get
$$\begin{gathered} \Rightarrow 2\left({\displaystyle \frac{1}{2}}{\log }_{5}11\right)=2y \\ \Rightarrow {\log }_{5}11=2y \end{gathered}$$
Let us now make use of the property of logarithm, $${\log }_{b}a={\displaystyle \frac{{\log }_{c}a}{{\log }_{c}b}}$$.
So, on applying this property in the above equation, we get
$$\Rightarrow {\displaystyle \frac{{\log }_{3}11}{{\log }_{3}5}}=2y$$
Substituting the value of $${\log }_{3}5$$ in the above expression, we get
$$\Rightarrow {\displaystyle \frac{{\log }_{3}11}{x}}=2y$$
Multiplying the above equation by $$x$$ on both sides, we get
$$\begin{gathered} \Rightarrow x\left({\displaystyle \frac{{\log }_{3}11}{x}}\right)=2xy \\ \Rightarrow {\log }_{3}11=2xy \end{gathered}$$
Rewriting the expression $${\log }_3{\displaystyle \frac{11}{3}}$$ using the logarithm property, $${\log }_c\left({\displaystyle \frac{a}{b}}\right)={\log }_{c}a-{\log }_{c}b$$, we get
$$\Rightarrow {\log }_3\left({\displaystyle \frac{11}{3}}\right)={\log }_{3}11-{\log }_{3}3$$
Substituting the values of $${\log }_{3}11$$ and $${\log }_{3}3$$ in the above expression, we get
$$\Rightarrow {\log }_3\left({\displaystyle \frac{11}{3}}\right)=2xy-1$$
Thus, the value of $${\log }_3{\displaystyle \frac{11}{3}}$$ is $$2xy-1$$.


Note: The logarithm rules can be used for fast exponent calculation using multiplication operation. Students should make use of the appropriate formula of logarithms wherever needed and solve the problem. In mathematics, if the base value in the logarithm function is not written, then the base is $$e$$.