Question

If x satisfies the equation \[{{\log }_{\dfrac{1}{3}}}\left( x-1 \right)+{{\log }_{\dfrac{1}{3}}}\left( x+1 \right)+{{\log }_{\sqrt{3}}}\left( 5-x \right)<1\], then x lies in the interval
(a) (1, 3)
(b) (2, 5)
(c) (1, 2)
(d) \[(4,\infty )\]

Answer 1

Hint: First of all find the value of x by using the domain of a logarithmic function that is if \[{{\log }_{a}}x\], x > 0, a > 0 and \[a\ne 1\]. Now, solve the given equation and substitute in it the values of x that you got earlier to find the final interval of x.

Complete step-by-step answer:

We are given that if x satisfies equation \[{{\log }_{\dfrac{1}{3}}}\left( x - 1 \right)+{{\log }_{\dfrac{1}{3}}}\left( x+1 \right)+{{\log }_{\sqrt{3}}}\left( 5-x \right)<1\] then we have to find the interval in which x lies. Let us consider the equation given in the question
\[{{\log }_{\dfrac{1}{3}}}\left( x-1 \right)+{{\log }_{\dfrac{1}{3}}}\left( x+1 \right)+{{\log }_{\sqrt{3}}}\left( 5-x \right)<1\]
First of all, we know that in \[{{\log }_{a}}x\], x and a must be greater than 0 and \[a\ne 1\]. So we get,
\[\left( x-1 \right)>0\]
So, \[x>1.....\left( i \right)\]
And \[\left( x+1 \right)>0\]
So, \[x>-1....\left( ii \right)\]
And, \[\left( 5-x \right)>0\]
\[x<5....\left( iii \right)\]
By considering equation (i), (ii) and (iii), we get,
\[x\in \left( 1,5 \right)....\left( iv \right)\]
Now, let us solve the given equation.
\[{{\log }_{\dfrac{1}{3}}}\left( x-1 \right)+{{\log }_{\dfrac{1}{3}}}\left( x+1 \right)+{{\log }_{\sqrt{3}}}\left( 5-x \right)<1\]
We know that \[\dfrac{1}{a}={{a}^{-1}}\] and \[\sqrt{a}={{a}^{\dfrac{1}{2}}}\]. By using this in the above equation, we get,
\[{{\log }_{{{\left( 3 \right)}^{-1}}}}\left( x-1 \right)+{{\log }_{{{\left( 3 \right)}^{-1}}}}\left( x+1 \right)+{{\log }_{{{\left( 3 \right)}^{\dfrac{1}{3}}}}}\left( 5-x \right)<1\]
We know that \[{{\log }_{{{a}^{n}}}}{{x}^{m}}=\dfrac{m}{n}{{\log }_{a}}x\]. By using this in the above equation, we get,
\[\dfrac{1}{-1}{{\log }_{3}}\left( x-1 \right)+\left( \dfrac{1}{-1} \right){{\log }_{3}}\left( x+1 \right)+\dfrac{1}{\dfrac{1}{3}}{{\log }_{3}}\left( 5-x \right)<1\]
\[\Rightarrow -{{\log }_{3}}\left( x-1 \right)-{{\log }_{3}}\left( x+1 \right)+3{{\log }_{3}}\left( 5-x \right)<1\]
We know that \[n{{\log }_{a}}m={{\log }_{a}}{{m}^{n}}\]. By using this in the above equation, we get,
\[{{\log }_{3}}{{\left( x-1 \right)}^{-1}}+{{\log }_{3}}{{\left( x+1 \right)}^{-1}}+{{\log }_{3}}{{\left( 5-x \right)}^{3}}<1\]
Now we know that \[{{\log }_{a}}{{m}_{1}}+{{\log }_{a}}{{m}_{2}}+{{\log }_{a}}{{m}_{3}}....={{\log }_{a}}{{m}_{1}}{{m}_{2}}{{m}_{3}}....\]
By using this in the above equation, we get,
\[{{\log }_{3}}{{\left( x-1 \right)}^{-1}}{{\left( x+1 \right)}^{-1}}{{\left( 5-x \right)}^{3}}<1\]
\[\Rightarrow {{\log }_{3}}\dfrac{{{\left( 5-x \right)}^{3}}}{\left( x-1 \right)\left( x+1 \right)}<1\]
We can also write \[{{\log }_{a}}m < b\] as \[m < {{\left( a \right)}^{b}}\] if \[a > 1\]. By using this in the above equation, we get,
\[\dfrac{{{\left( 5-x \right)}^{3}}}{\left( x-1 \right)\left( x+1 \right)}<{{\left( 3 \right)}^{1}}\]
\[\dfrac{{{\left( 5-x \right)}^{3}}}{\left( x-1 \right)\left( x+1 \right)}<3....\left( v \right)\]
Now as we have already calculated that \[x\in \left( 1,5 \right)\], we can find the values of x from the above equation by substituting the values in the small intervals between (1, 5) and checking if it is satisfying the above equation or not.
So, let us take \[x\in \left( 1,2 \right)\] for instance take x = 1.5
By substituting x = 1.5 in equation (v), we get,
\[\dfrac{{{\left( 5-1.5 \right)}^{3}}}{\left( 1.5-1 \right)\left( 1.5+1 \right)}<3\]
\[\dfrac{\left( 42.875 \right)}{\left( 0.5 \right)\left( 2.5 \right)}<3\]
\[34.3<3\]
So, we can see that the above inequality is not true. So \[x\notin \left( 1,2 \right)\].
Now, let us take \[x\in \left( 2,3 \right)\] for instance take x = 2.5
By substituting x = 2.5 in equation (v), we get,
\[\dfrac{{{\left( 5-2.5 \right)}^{3}}}{\left( 2.5-1 \right)\left( 2.5+1 \right)}<3\]
\[\dfrac{15.625}{\left( 1.5 \right)\left( 3.5 \right)}<3\]
\[2.97<3\]
As we can see that the above inequality is true. So, \[x\in \left( 2,3 \right)\].
Now, let us take \[x\in \left( 3,4 \right)\] for instance take x = 3.5
By substituting x = 3.5 in equation (v), we get,
\[\dfrac{{{\left( 5-3.5 \right)}^{3}}}{\left( 3.5-1 \right)\left( 3.5+1 \right)}<3\]
\[\dfrac{3.375}{\left( 2.5 \right)\left( 4.5 \right)}<3\]
\[0.3<3\]
As we can see that the above inequality is true. So, \[x\in \left( 3,4 \right)\].
Now, let us take \[x\in \left( 4,5 \right)\] for instance take x = 4.5
By substituting x = 4.5 in equation (v), we get,
\[\dfrac{{{\left( 5-4.5 \right)}^{3}}}{\left( 4.5-1 \right)\left( 4.5+1 \right)}<3\]
\[\dfrac{0.125}{\left( 3.5 \right)\left( 5.5 \right)}<3\]
\[0.00649<3\]
As we can see that the above inequality is true. So, \[x\in \left( 4,5 \right)\].
So, \[x\in \left( 2,3 \right)\cup \left( 3,4 \right)\cup \left( 4,5 \right)\].
So, we finally get \[x\in \left( 2,5 \right)\].
Hence, option (b) is the right answer.

Note: In this question, first of all, it is very important to check the domain of the function that is for \[{{\log }_{a}}x,x>0,a>0\] and \[a\ne 1\]. Now for the equation \[\dfrac{{{\left( 5-x \right)}^{3}}}{\left( x-1 \right)\left( x+1 \right)}<3\], it is better to substitute x in the interval (1, 5) than finding x by solving the equation because it will become very long if we go for solving the equation.