Question

If the terms $\log x$, $\log \sqrt {6 - 2x}$, $\log \left( {x - 1} \right)$ are in an A.P. then, what are the number of possible values of $x$?
A) $0$
B) $1$
C) $2$
D) $3$

Answer 1

In this question, we are given three terms in log and we have been told that they are in A.P. We have been asked to find the value of x. Using the log property $\log c$, $c > 0$, make the conditions and try to find a number between those conditions. You can also use the common difference of A.P to find the required value. Keep the terms in the expression $2b = a + c$ and use log properties to simplify the equation. On simplifying, you will get the value of x.

Complete step-by-step solution:
We are given 3 terms in log and we have been told that they are in an A.P. We have been asked to find the value of $x$.
We will solve this question using 2 following conditions –

1. In $\log c$, $c > 0$. So, we will use this condition to determine the possible values.
2. Difference between the terms in an A.P is the same.
   $\Rightarrow$ Using condition 1, we can say that –
   1)$x > 0$,
   2)$6 - 2x > 0$
   $\Rightarrow 6 > 2x$
   $\Rightarrow 3 > x$
   3)$x - 1 > 0$
   $\Rightarrow x > 1$
   Therefore, our $x$ is greater than 0 and 1 and smaller than 3. Basically, it lies between 1 and 3. Hence, the value of $x$ is 2. But, we will cross verify this with condition 2.
   $\Rightarrow$ In 2nd condition, we will use the given fact that the three given terms are in A.P.
   If a, b, c are in an A.P, then $2b = a + c$. We will use this on the given terms.
   $\log x$, $\log \sqrt {6 - 2x}$, $\log \left( {x - 1} \right)$ are in A.P.
   $\Rightarrow 2\log \sqrt {6 - 2x} = \log x + \log \left( {x - 1} \right)$
   Using properties of log to solve the equation, $\left( {2\log x = \log {x^2}{\text{ and }}\log m + \log n = \log mn} \right)$
   $\Rightarrow \log {\left( {\sqrt {6 - 2x} } \right)^2} = \log x\left( {x - 1} \right)$
   Comparing both the sides,
   $\Rightarrow {\left( {\sqrt {6 - 2x} } \right)^2} = x\left( {x - 1} \right)$
   Simplifying the equation,
   $\Rightarrow 6 - 2x = {x^2} - x$
   $\Rightarrow {x^2} + x - 6 = 0$
   Now, we have a quadratic equation. Let us solve it –
   $\Rightarrow {x^2} + 3x - 2x - 6 = 0$
   $\Rightarrow \left( {x + 3} \right)\left( {x - 2} \right) = 0$
   This gives us two values of x,
   $\Rightarrow x = - 3,2$
   But x cannot be negative because there is a rule of log which says that – in $\log c$, $c > 0$.
   Therefore, the required value of x is option C) 2.

Note: We have remembered that, in mathematics, the logarithm is the inverse function to exponentiation. That means the logarithm of a given number x is the exponent to which another fixed number, the base b, must be raised, to produce that number x. In the simplest case, the logarithm counts the number of occurrences of the same factor in repeated multiplication.