Question

How do you solve ${\log _3}x = 9{\log _x}3$?

Answer 1

Bring like terms on the same side and use log properties to solve the equation To solve this question, the first thing we will have to do is bring all the like terms aside. After doing this we use a log property which states that $\dfrac{{\log a}}{{\log b}} = \log (a - b)$. After this we will use another log property on the resulting part which states that ${\log _b}a = \dfrac{{{{\log }_z}a}}{{{{\log }_z}b}}$. Basically, this step is to simplify the question and get our answer. You will need to practice to get ideas like this. Here, we will divide the log in such a way that we get a much solvable version of the question. Post this step; we will use another property which states that ${\log _b}a = c \to {b^c} = a$. After this, we will solve for x which will in the end give us our answer.

Complete step by step solution:
The given question we have is ${\log _3}x = 9{\log _x}3$
To solve this, we will use a special property of log which states that:-
$\dfrac{{\log a}}{{\log b}} = \log (a - b)$
Now, beginning to solve the above problem, we will take terms on the same side of the equation.
So, when we take logs on one side, we will get:-
$\dfrac{{{{\log }_3}x}}{{{{\log }_x}3}} = 9$
At this point, we will use another property of log which states that
${\log _b}a = \dfrac{{{{\log }_z}a}}{{{{\log }_z}b}}$
Using the above formula, we will get the modified version of our question as:-
$\dfrac{{\dfrac{{\log x}}{{\log 3}}}}{{\dfrac{{\log 3}}{{\log x}}}} = 9 \\\ \Rightarrow \dfrac{{\log x}}{{\log 3}} \times \dfrac{{\log x}}{{\log 3}} = 9 \\\ \Rightarrow {\left( {\dfrac{{\log x}}{{\log 3}}} \right)^2} = 9 \\\$
Using the very first formula we mentioned, our equation will be
${\left( {\log (x - 3)} \right)^2} = 9$
Taking square root on both the sides, we will get:-
\sqrt {{{\left( {\log (x - 3)} \right)}^2}} = \sqrt 9 \\\ \Rightarrow \log (x - 3) = 3 \\\
Now, since we have taken the normal log. The base of this equation will be 10.
At this point, to simplify this further. We will use another formula, we will get:-
${\log _b}a = c \to {b^c} = a$
$\log (x - 3) = 3 \\\ \Rightarrow {10^3} = x - 3 \\\$
Rearranging the terms, we will get
$x - 3 = 1000 \\\ \Rightarrow x = 1000 - 3 \\\ \Rightarrow x = 1997 \\\$
Therefore, 1997 is the required value of x.

Note: Usage of log properties should be done when you have addition/multiplication/division/ /subtraction of the logs. These operations are not done in the conventional manner. You cannot do the conventional operation to the log arguments. Doing this will give you the wrong answer.