Question

If ${\log _3}2$, ${\log _3}\left( {{2^x} - 5} \right)$, ${\log _3}\left( {{2^x} - \dfrac{7}{2}} \right)$ are in AP. Then find the value of $x$.
A. 2
B. 3
C. 4
D. 2,3

Answer 1

Hint: Given that ${\log _3}2$, ${\log _3}\left( {{2^x} - 5} \right)$, ${\log _3}\left( {{2^x} - \dfrac{7}{2}} \right)$ are AP, we will apply the condition of the AP terms. Then solve the equation for x. After that put the solutions of x in the given terms to check which solution satisfies the terms.

Formula Used:
If $a,b,c$ are in Ap then $2b = a + c$.
$n\log m = \log {m^n}$
$\log a + \log b = \log ab$
${\log _a}b = {\log _a}c \Rightarrow b = c$
${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$
${a^x} = {a^y} \Rightarrow x = y$

Complete step by step solution:
Given that, ${\log _3}2$, ${\log _3}\left( {{2^x} - 5} \right)$, ${\log _3}\left( {{2^x} - \dfrac{7}{2}} \right)$ are AP.
Now we will apply the formula $2b = a + c$ where $a,b,c$ are in Ap.
Here $a = {\log _3}2$, $b = {\log _3}\left( {{2^x} - 5} \right)$ and $c = {\log _3}\left( {{2^x} - \dfrac{7}{2}} \right)$
Therefore,
$2{\log _3}\left( {{2^x} - 5} \right) = {\log _3}2 + {\log _3}\left( {{2^x} - \dfrac{7}{2}} \right)$
Now we will apply $n\log m = \log {m^n}$ on left side of the equation
${\log _3}{\left( {{2^x} - 5} \right)^2} = {\log _3}2 + {\log _3}\left( {{2^x} - \dfrac{7}{2}} \right)$
Now we will apply $\log a + \log b = \log ab$ on the right side of the equation
${\log _3}{\left( {{2^x} - 5} \right)^2} = {\log _3}2\left( {{2^x} - \dfrac{7}{2}} \right)$
Then apply the formula ${\log _a}b = {\log _a}c \Rightarrow b = c$
${\left( {{2^x} - 5} \right)^2} = 2\left( {{2^x} - \dfrac{7}{2}} \right)$
Now we will apply the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$
 ${2^{2x}} - 2 \cdot 5 \cdot {2^x} + {5^2} = 2\left( {{2^x} - \dfrac{7}{2}} \right)$
$ \Rightarrow {2^{2x}} - 10 \cdot {2^x} + 25 = 2 \cdot {2^x} - 7$
$ \Rightarrow {2^{2x}} - 12 \cdot {2^x} + 32 = 0$
Now apply the factorization method
${2^{2x}} - 8 \cdot {2^x} - 4 \cdot {2^x} + 32 = 0$
$ \Rightarrow {2^x}\left( {{2^x} - 8} \right) - 4\left( {{2^x} - 8} \right) = 0$
$ \Rightarrow \left( {{2^x} - 8} \right)\left( {{2^x} - 4} \right) = 0$
Equate each factor with zero
${2^x} - 8 = 0$ or, ${2^x} - 4 = 0$
$ \Rightarrow {2^x} = 8$ $ \Rightarrow {2^x} = 4$
$ \Rightarrow {2^x} = {2^3}$ $ \Rightarrow {2^x} = {2^2}$
$ \Rightarrow x = 3$ $ \Rightarrow x = 2$
Now we put $x = 2$ in ${\log _3}\left( {{2^x} - 5} \right)$.
${\log _3}\left( {{2^2} - 5} \right) = {\log _3}\left( { - 1} \right)$
Since the logarithm of a negative does not exist. So $x \ne 2$.
Therefore $x = 3$.

Option ‘B’ is correct

Note: If three terms are in AP then the difference between consecutive terms is constant. To solve the question you need to use the above concept to make an equation. Remember all solutions of $x$ might not satisfy the logarithmic terms, since the logarithmic of a negative number is undefined.