Question

Solve the equation:
\[{\log _9}81 + \dfrac{{{{\log }_9}1}}{9} + {\log _9}3 = {\log _9}x\]

Answer 1

Hint: To evaluate the equation, we need to solve LHS and then comparing L.H.S. and R.H.S, we can conclude to the solution. Also, it is very important for us to know the properties of logarithm as:-
• \[{\log _a}{x^n} = n{\log _a}x\]
• \[{\log _a}1 = 0\]
• \[{\log _a}{a^r} = r\]

Complete step by step solution:
To solve the equation, first we need to simplify L.H.S.
Therefore, we get; L.H.S. \[ = {\log _9}81 + \dfrac{{{{\log }_9}1}}{9} + {\log _9}3\]
Now, as we know that \[81 = {9^2}\] , Therefore, we get :-
\[\Rightarrow\] L.H.S. \[ = {\log _9}({9^2}) + \dfrac{{{{\log }_9}1}}{9} + {\log _9}3\]
\[\Rightarrow\] LHS \[ = {\log _9}{9^2} + \dfrac{{{{\log }_9}1}}{9} + {\log _9}3\]
According to logarithmic property, \[{\log _a}{a^r} = r\], we get;
\[\Rightarrow\] LHS \[ = 2 + \dfrac{{{{\log }_9}1}}{9} + {\log _9}3\] ( we simply apply the above given property )
Now, by using another logarithmic property, \[{\log _a}1 = 0\] ( where a can be any any positive number except 1 )
We get ; L.H.S. \[ = 2 + \dfrac{0}{9} + {\log _9}3\]
Also, we can rewrite \[3\]as :- \[3 = \sqrt 9 \,\,\,\, = \,\,{9^{\dfrac{1}{2}}}\]( for simplifying the expression ), we get;
\[\Rightarrow\] L.H.S. \[ = 2 + 0 + {\log _9}({9^{\dfrac{1}{2}}})\]
Again, by property of logarithm, \[{\log _a}{a^r} = \,r\], we get;
\[\Rightarrow\] L.H.S. \[ = 2 + 0 + \dfrac{1}{2}\]
\[\Rightarrow\therefore \]L.H.S. \[ = 2 + \dfrac{1}{2}\]
\[\Rightarrow = \dfrac{5}{2}\] ( by taking L.C.M )
Thus, we get
\[\Rightarrow{\log _9}x = \dfrac{5}{2}\]
By rewriting it in exponential form, we get;
\[\Rightarrow x = {9^{\dfrac{5}{2}}}\] ( we apply the most basic property of log which is given above )
\[ = {3^5}\] ( root of 9 is 3 )
\[ = 243\], which is our required answer.

Note: Such logarithmic problem where base of L.H.S. and R.H.S. are some can be solved by alternative way by using logarithmic ‘Product rule’ which states:-
\[{\log _b}(MN) = {\log _b}(M) + {\log _b}(N)\]
And quotient rule which states:-
\[\Rightarrow{\log _b}\left( {\dfrac{M}{N}} \right) = {\log _b}(M) - {\log _b}(N)\]
\[\Rightarrow\therefore \]L.H.S. \[ = {\log _9}81 + \dfrac{{{{\log }_9}1}}{9} + {\log _9}3\]
Now, As we know, \[{\log _a}1 = 0\]
\[\Rightarrow \therefore \] L.H.S. \[{\log _9}81 + 0 + {\log _9}3\,\,\, = \,\,\,{\log _9}81 + {\log _9}3\]
Now, By applying the product rule, we get;
L.H.S. \[ = {\log _9}(81 \times 3)\,\,\,\,\,\,\,\, = \,lo{g_9}(243)\]
and we have, R.H.S. = \[{\log _9}x\]
By comparing L.H.S. & R.H.S, we get;
\[{\log _9}x = {\log _9}(243)\]
\[x = 243\].