Question

If the terms ${{\log }_{3}}2,{{\log }_{3}}\left( {{2}^{x}}-5 \right)$ and ${{\log }_{3}}\left( {{2}^{x}}-\dfrac{7}{2} \right)$ are in A.P, then $x$ is equal to:
1. 2
2. 3
3. 4
4. 2, 3

Answer 1

Hint: For solving this question you should know about the arithmetic progression. As we know that the terms of any A.P are as that the double of any term is the sum of it’s both left side and right side of the terms. And it can be written as $2x=y+z$, if the middle term is $x$ and the left side term and right side terms are $y$ and $z$ respectively. Here we will use this property.

Complete step-by-step solution:
According to the question it is asked to us to find the value of $x$, if ${{\log }_{3}}2,{{\log }_{3}}\left( {{2}^{x}}-5 \right)$ and ${{\log }_{3}}\left( {{2}^{x}}-\dfrac{7}{2} \right)$ are in A.P. As it is given that the following terms are in A.P respectively. So, they are,
 ${{\log }_{3}}2,{{\log }_{3}}\left( {{2}^{x}}-5 \right),{{\log }_{3}}\left( {{2}^{x}}-\dfrac{7}{2} \right)$
SO, it can be written by the property of A.P, $2x=y+z$ as follows,
$\Rightarrow {{\log }_{3}}2+{{\log }_{3}}\left( {{2}^{x}}-\dfrac{7}{2} \right)=2\times {{\log }_{3}}\left( {{2}^{x}}-5 \right)$
Now, if we solve this equation, we will get as follows,
$\begin{align}
  & \Rightarrow {{\log }_{3}}\left( 2\times \left( {{2}^{x}}-\dfrac{7}{2} \right) \right)={{\log }_{3}}{{\left( {{2}^{x}}-5 \right)}^{2}} \\
 & \because {{\log }_{x}}\left( a \right)+{{\log }_{x}}\left( b \right)={{\log }_{x}}\left( a\times b \right) \\
\end{align}$
Now, by taking anti-log on both the sides, we will get,
$\begin{align}
  & \Rightarrow {{2}^{x+1}}-7={{\left( {{2}^{x}}-5 \right)}^{2}} \\
 & \Rightarrow {{2}^{x+1}}-7={{2}^{2x}}-5\times {{2}^{x+1}}+25 \\
\end{align}$
Now we will take all the terms of the left hand side. Therefore we will get,
$\begin{align}
  & \Rightarrow {{2}^{2x}}-6\times {{2}^{x+1}}+32=0 \\
 & \Rightarrow {{2}^{2x}}-12\times {{2}^{x}}+32=0 \\
 & \Rightarrow \left( {{2}^{x}}-4 \right)\times \left( {{2}^{x}}-8 \right)=0 \\
 & \Rightarrow {{2}^{x}}=4,8\text{ or }x=2,3 \\
\end{align}$
However when $x$ is 2, ${{2}^{x}}-5$ becomes -1 which is not allowed inside the logarithm. So, the value of $x$ will be $x=3$.
Hence the correct answer is option 2.

Note: While solving any question for any logarithmic term, you have to ensure that all the logarithmic properties must be clear to you because they can only be solved by using these properties, otherwise they can’t be solved.