Question

If $ \log 2,\log \left( {{2}^{x}}-1 \right),\log \left( {{2}^{x}}+3 \right) $ are three consecutive terms of an AP, then the value of x will be
a. $ {{\log }_{2}}5 $
b. 0
c. $ {{\log }_{5}}2 $
d. None of these

Answer 1

Hint: As we are given that the three terms $ \log 2,\log \left( {{2}^{x}}-1 \right),\log \left( {{2}^{x}}+3 \right) $ are in AP, we can write, $ \log \left( {{2}^{x}}-1 \right)-\log 2=\log \left( {{2}^{x}}+3 \right)-\log \left( {{2}^{x}}-1 \right) $ . We will also use the formula, $ \log a-\log b=\log \left( \dfrac{a}{b} \right) $ . By using these relations, we will find the value of x.

Complete step-by-step answer:
It is given in the question that $ \log 2,\log \left( {{2}^{x}}-1 \right),\log \left( {{2}^{x}}+3 \right) $ are three consecutive terms of an AP, and we have been asked to find the value of x. We know that in an AP, there will be a common difference between any two consecutive terms, so for the given terms we can write as,
 $ \log \left( {{2}^{x}}-1 \right)-\log 2=\log \left( {{2}^{x}}+3 \right)-\log \left( {{2}^{x}}-1 \right) $
Now, we know that $ \log a-\log b=\log \left( \dfrac{a}{b} \right) $ . So, we can write,
  $ \log \left( \dfrac{{{2}^{x}}-1}{2} \right)=\log \left( \dfrac{{{2}^{x}}+3}{{{2}^{x}}-1} \right) $
On cancelling the log on both the sides, we get,
\[\left( \dfrac{{{2}^{x}}-1}{2} \right)=\left( \dfrac{{{2}^{x}}+3}{{{2}^{x}}-1} \right)\]
Now, we will assume $ \left( {{2}^{x}} \right)=y $ . So, we get,
\[\dfrac{\left( y-1 \right)}{2}=\dfrac{\left( y+3 \right)}{\left( y-1 \right)}\]
On cross multiplying both sides of the above expression, we get,
\[\begin{align}
  & \left( y-1 \right)\left( y-1 \right)=2\left( y+3 \right) \\
 & {{y}^{2}}-y-y+1=2y+6 \\
 & {{y}^{2}}-2y+1=2y+6 \\
\end{align}\]
On transposing 2y and 6 from the RHS to the LHS, we get,
 $ \begin{align}
  & {{y}^{2}}-2y-2y+1-6=0 \\
 & {{y}^{2}}-4y-5=0 \\
\end{align} $
On splitting the term -4y as -5y + y, we get,
 $ \begin{align}
  & {{y}^{2}}-5y+y-5=0 \\
 & y\left( y-5 \right)+\left( y-5 \right)=0 \\
 & \left( y+1 \right)\left( y-5 \right)=0 \\
 & y=-1,5 \\
\end{align} $
Now, y cannot be equal to -1 because an exponential function $ {{f}^{n}} $ is not negative so, -1 is neglected. So, we get,
y = 5
Now, we know that $ y=\left( {{2}^{x}} \right) $ , so we get,
 $ {{2}^{x}}=5 $
On taking log on both the sides, we get,
 $ \log {{2}^{x}}=\log 5 $
Now, we know that $ \log {{a}^{b}} $ can be written as $ b\log a $ , so we get,
 $ \begin{align}
  & x\log 2=\log 5 \\
 & x=\dfrac{\log 5}{\log 2} \\
 & x={{\log }_{2}}5 \\
\end{align} $
Thus, option (a), $ {{\log }_{2}}5 $ is the correct answer.

Note:While solving this question, most of the students directly cross multiply \[\left( \dfrac{{{2}^{x}}-1}{2} \right)=\left( \dfrac{{{2}^{x}}+3}{{{2}^{x}}-1} \right)\] as a result they might get a complicated answer because of the exponential powers. Thus it is recommended to assume some variable for the exponential term, \[{{2}^{x}}\] in order to make the equation simple and the calculations easier. Also, in the last step the students must remember to substitute the value of the variable as \[{{2}^{x}}\]back in the final equation.