Question

If a, b $\ne 1$, ab $>0$, a ≠ b and ${{\log }_{b}}a={{\log }_{a}}b$, then ab = ?
(a) $\dfrac{1}{2}$
(b) $1$
(c) $2$
(d) $10$

Answer 1

Hint : To solve this question, apply the logarithm rules to arrive at a simpler equation which can be solved while keeping in mind the conditions given in the question. The definition of logarithm is very important and most of the questions involving logarithms can be solved using its definition.

Complete step by step solution:
${{\log }_{b}}x$ means “to what exponent we must raise b in order to get x”.
Look at this example to understand the above definition:
${{\log }_{2}}8=3$ because ${{2}^{3}}=8$
Therefore, a logarithmic function is the inverse of an exponential function.
Mathematically,
${{\log }_{b}}x=a$ if ${{b}^{a}}=x$
$b>0,\ne 1,$ and $x>0$
It is read as log base b of x is a
Now, Let’s solve the question using some properties of log. Let’s start with what is given to us.
${{\log }_{b}}a={{\log }_{a}}b$
Now using the change of base rule property, which states that:
${{\log }_{a}}b=\dfrac{\log b}{\log a}$
Remember, when the base of the log is not mentioned it is assumed to be $10$
Using the above property, we get
$\dfrac{\log a}{\log b}=\dfrac{\log b}{\log a}$

$$\Rightarrow {{(\log a)}^{2}}={{(\log b)}^{2}} \\\ \Rightarrow {{(\log a)}^{2}}-{{(\log b)}^{2}}=0 \\\ \Rightarrow (\log a-\log b)(\log a+\log b)=0 \\\$$

Either $\log a-\log b=0$ or $\log a+\log b=0$
Now, applying the division and product rule on the two logarithmic equations and writing$0$as $\log 1$.
$\Rightarrow \log \dfrac{a}{b}=\log 1$ or $\log ab=\log 1$
Comparing the both sides of the logarithmic equation
$\Rightarrow \dfrac{a}{b}=1$ or $ab=1$
$\Rightarrow a=b$ or $ab=1$
But is given that a ≠ b
$\therefore ab=1$
Hence, option ‘b’ is the correct option.
So, the correct answer is “Option B”.

Note : While solving questions like these, we should always start with what is given to us and step by step reduce the given equation to a simpler one(here we broke it down to a simple algebraic equation). Usually, students apply the wrong rules to log in these questions.