Question

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

Answer 1

Hint: Here, we will use the logarithm properties like, \[{\log _{{b^2}}}a = \dfrac{1}{2}{\log _b}a\] , \[{\log _b}a = \dfrac{{{{\log }_c}a}}{{{{\log }_c}b}}\] and \[{\log _c}\left( {\dfrac{a}{b}} \right) = {\log _c}a - {\log _c}b\] 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, \[{\log _{{b^2}}}a = \dfrac{1}{2}{\log _b}a\] in the above expression, we get
\[ \Rightarrow \dfrac{1}{2}{\log _5}11 = y\]
Multiplying the above equation by 2 on both sides, we get
\[
   \Rightarrow 2\left( {\dfrac{1}{2}{{\log }_5}11} \right) = 2y \\
   \Rightarrow {\log _5}11 = 2y \\
 \]
Let us now make use of the property of logarithm, \[{\log _b}a = \dfrac{{{{\log }_c}a}}{{{{\log }_c}b}}\].
So, on applying this property in the above equation, we get
\[ \Rightarrow \dfrac{{{{\log }_3}11}}{{{{\log }_3}5}} = 2y\]
Substituting the value of \[{\log _3}5\] in the above expression, we get
\[ \Rightarrow \dfrac{{{{\log }_3}11}}{x} = 2y\]
Multiplying the above equation by \[x\] on both sides, we get
\[
   \Rightarrow x\left( {\dfrac{{{{\log }_3}11}}{x}} \right) = 2xy \\
   \Rightarrow {\log _3}11 = 2xy \\
 \]
Rewriting the expression \[{\log _3}\dfrac{{11}}{3}\] using the logarithm property, \[{\log _c}\left( {\dfrac{a}{b}} \right) = {\log _c}a - {\log _c}b\], we get
\[ \Rightarrow {\log _3}\left( {\dfrac{{11}}{3}} \right) = {\log _3}11 - {\log _3}3\]
Substituting the values of \[{\log _3}11\] and \[{\log _3}3\] in the above expression, we get
\[ \Rightarrow {\log _3}\left( {\dfrac{{11}}{3}} \right) = 2xy - 1\]
Thus, the value of \[{\log _3}\dfrac{{11}}{3}\] 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 \[e\].