Question

If ${\log }_{3}2,{\log }_3(2x-5)$ and ${\log }_3(2^x-\frac{7}{2})$ are in AP, then $x$ is equal to

• A. $8$
• B. $3$
• C. $-8$
• D. $-3$

Answer 1

The correct option is B

$3$

Explanation for the correct answer:

Step 1: Simplify the given equation using log properties

Given: ${\log }_{3}2,{\log }_3(2x-5)$ and ${\log }_3(2^x-\frac{7}{2})$ are in AP

If $a,b,c$ are in A.P then $2b=a+c$

$\begin{array}{crrll}\Rightarrow & 2{\log }_3\left(2^x-5\right)& =& {\log }_3\left(2\right)+{\log }_3\left(2^x-\frac{7}{2}\right)& \\ \Rightarrow & {\log }_3{\left(2^x-5\right)}^2& =& {\log }_3\left[2\left(2^x-\frac{7}{2}\right)\right]& \left[\because \mathrm{alog}\left(b\right)=\log {\left(b\right)}^a;\log \left(a\right)+\log \left(b\right)=\log \left(\mathrm{ab}\right)\right]\\ \Rightarrow & {\left(2^x-5\right)}^2& =& 2\left(2^x-\frac{7}{2}\right)& \\ \Rightarrow & 2^{2x}+25-10·2^x& =& 2·2^x-7& \left[\because {\left(a-b\right)}^2=a^2+b^2-2\mathrm{ab}\right]\\ \Rightarrow & 2^{2x}-12·2^x+32& =& 0& \end{array}$

Step 2: Factorize the equation to get the value of $x$

Let $2^x=y$

$\begin{array}{crcl}\Rightarrow & y^2-12y+32& =& 0\\ \Rightarrow & y^2-8y-4y+32& =& 0\\ \Rightarrow & y(y-8)-4(y-8)& =& 0\\ \Rightarrow & (y-4)(y-8)& =& 0\\ \Rightarrow & y& =& 4,8\end{array}$

$$\begin{gathered} \therefore 2^x=4or{2}^x=8 \\ \Rightarrow x=2orx=3 \end{gathered}$$ $\left[\because \mathrm{at}x=2{\log }_3\left(2x-5\right)\mathrm{will}\mathrm{fail}\right]$

Hence option (B) i.e. $3$ is correct.