Question

If $ {\log _b}a = \dfrac{1}{x} $ and $ {\log _a}\sqrt b = 3{x^2} $ , show that $ x = \dfrac{1}{6} $ ?

Answer 1

Hint: Here, in both the given terms, we can see that the bases are different. Therefore we have to apply the change of base formula for both the terms. We will also apply the exponent rule for logarithms in the second term. By doing this, we will get the value of $ x $ .
Formulas used:
Change of base rule for logarithms: $ {\log _q}p = \dfrac{{{{\log }_e}p}}{{{{\log }_e}q}} $
Exponent rule for logarithms: $ \log {p^q} = q\log p $

Complete step-by-step answer:
Let us first consider the first term $ {\log _b}a = \dfrac{1}{x} $ .
We will apply the change of base formula $ {\log _q}p = \dfrac{{{{\log }_e}p}}{{{{\log }_e}q}} $ here for the left hand side.
\[{\log _b}a = \dfrac{{{{\log }_e}a}}{{{{\log }_e}b}}\]
Therefore, our first term will be \[\dfrac{{{{\log }_e}a}}{{{{\log }_e}b}} = \dfrac{1}{x}\].
Now, let us consider the second term $ {\log _a}\sqrt b = 3{x^2} $ .
Here we can rewrite the left hand side as $ {\log _a}{b^{\dfrac{1}{2}}} $ . We will now apply the exponent rule $ \log {p^q} = q\log p $ , and thus we can say that $ {\log _a}{b^{\dfrac{1}{2}}} = \dfrac{1}{2}{\log _a}b $ .
Therefore, our second term will be $ \dfrac{1}{2}{\log _a}b = 3{x^2} $ .
If we again apply the change of base formula, we get \[{\log _a}b = \dfrac{{{{\log }_e}b}}{{{{\log }_e}a}}\].
Therefore, the second term can be written as \[\dfrac{1}{2}\left( {\dfrac{{{{\log }_e}b}}{{{{\log }_e}a}}} \right) = 3{x^2}\].
We have determined the first term as
\[
  \dfrac{{{{\log }_e}a}}{{{{\log }_e}b}} = \dfrac{1}{x} \\
   \Rightarrow \dfrac{{{{\log }_e}b}}{{{{\log }_e}a}} = x \;
 \]
We will now put this value in the second term.
\[
  \dfrac{1}{2}\left( {\dfrac{{{{\log }_e}b}}{{{{\log }_e}a}}} \right) = 3{x^2} \\
   \Rightarrow \dfrac{1}{2}x = 3{x^2} \\
   \Rightarrow \dfrac{1}{2} = 3x \\
   \Rightarrow x = \dfrac{1}{6} \;
 \]
Hence, it is proved that $ x = \dfrac{1}{6} $ .
So, the correct answer is “ $ x = \dfrac{1}{6} $ ”.

Note: In this type of question where a non-standard-base log is given, we need to keep in mind to use the change of base formula. We can evaluate a non-standard-base log by converting it to the fraction of the form standard-base log of the argument divided by the same-standard-base log of the non-standard-base. Also, it does not matter which standard-base log we use, as long as we use the same base for both the numerator and the denominator.