Question

If the terms \[1,{{\log }_{3}}\sqrt{{{3}^{1-x}}+2},{{\log }_{3}}({{4.3}^{x}}-1)\] are in AP. Then, x is equal to
(a) \[1 - {\log _3}4\]
(b) \[{\log _3}4\]
(c) \[{\log _3}7\]
(d) \[{\log _7}4\]

Answer 1

Hint: To deal with the problems regarding logarithm, we should try to make the bases of the log just the same. To make that we will use \[{{\log }_{3}}3=1\] so that we can use the rules of logarithm to go ahead with the problem. In such a way the solution of any logarithmic function can be reached easily.

Complete step by step solution:
We are given, \[1,{{\log }_{3}}\sqrt{{{3}^{1-x}}+2},{{\log }_{3}}({{4.3}^{x}}-1)\] are in arithmetic progression.
If three numbers a,b and c are in A.P then, 2b = a + c,
So, we get now,
\[2{\log _3}\sqrt {{3^{1 - x}} + 2} = 1 + {\log _3}({4.3^x} - 1)\]
\[\Rightarrow 2{\log _3}\sqrt {{3^{1 - x}} + 2} = {\log _3}3 + {\log _3}({4.3^x} - 1)\]
 as, \[{\log _3}3\] = 1
\[\Rightarrow 2{\log _3}{({3^{1 - x}} + 2)^{\dfrac{1}{2}}} = {\log _3}3 + {\log _3}({4.3^x} - 1)\]
\[\Rightarrow {\log _3}{({3^{1 - x}} + 2)^{\dfrac{1}{2}.2}} = {\log _3}3({4.3^x} - 1)\]
 As, [log a + log b = log ab]
\[\Rightarrow {\log _3}({3^{1 - x}} + 2) = {\log _3}({12.3^x} - 3)\]
\[\Rightarrow {3^{1 - x}} + 2 = {12.3^x} - 3\]
 As, log a = log b implies a = b
Now, let,
\[{3^x} = t\]
\[\Rightarrow \dfrac{3}{t} + 2 = 12t - 3\]
Now, we will take LCM. Therefore, we get
\[\begin{array}{l}
 \Rightarrow \dfrac{{3 + 2t}}{t} = 12t - 3\\
 \Rightarrow 3 + 2t = 12{t^2} - 3t\\
 \Rightarrow 12{t^2} - 5t - 3 = 0
\end{array}\]
And we know that 5t = 9t - 4t. Therefore, we get
\[\Rightarrow 12{t^2} - 9t + 4t - 3 = 0\]
\[\Rightarrow 3t(4t - 3) + 1(4t - 3) = 0\]
\[\Rightarrow (3t + 1)(4t - 3) = 0\]
\[\Rightarrow t = - \dfrac{1}{3},\dfrac{3}{4}\]
Now, putting the value,
\[\Rightarrow {3^x} = \dfrac{3}{4}\]
 As, \[{3^x}\] can’t be negative, we can’t choose our value to be negative. That is why only positive value has to be chosen.
\[\Rightarrow x = {\log _3}\dfrac{3}{4}\]
\[\Rightarrow x = {\log _3}3 - {\log _3}4\]
\[\Rightarrow x = 1 - {\log _3}4\]
Hence, the correct option is, (a) \[1 - {\log _3}4\]

Note: To solve the quadratic equation we can use the formula of Sridharacharya also. The formula is said to be, when, $a{{x}^{2}}+bx+c=0$ we get, $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .On the other hand, if we choose our value of t to be negative, then as per the rules of integration the value of x would be undefined. Analyzing the problem, in this way we will find our desired solution of x with the form of logarithmic function.