Question

If ${\log }_{10}2,{\log }_{10}(2^x–1)\text{and}{\log }_{10}(2^x+3)$ are in AP, then $x$ is equal to:

• A. ${\log }_{2}5$
• B. ${\log }_2\left(-1\right)$
• C. ${\log }_2\left(\frac{1}{5}\right)$
• D. ${\log }_{5}2$

Answer 1

The correct option is A

${\log }_{2}5$

Explanation for the correct option.

As ${\log }_{10}2,{\log }_{10}(2^x–1)\text{and}{\log }_{10}(2^x+3)$ are in AP, so

$\begin{array}{rcl}2{\log }_{10}(2^x–1)& =& {\log }_{10}2+{\log }_{10}(2^x+3)\\ \Rightarrow {(2^x–1)}^2& =& 2(2^x+3).....\left(1\right)\end{array}$

Let $2^x=y$, so the equation $\left(1\right)$, will be

$\begin{array}{rcl}{\left(y-1\right)}^2& =& 2\left(y+3\right)\\ \Rightarrow y^2-2y+1& =& 2y+6\\ \Rightarrow y^2-4y-5& =& 0\\ \Rightarrow y^2+y-5y-5& =& 0\\ \Rightarrow \left(y-5\right)\left(y+1\right)& =& 0\\ \Rightarrow y& =& 5\mathrm{or}-1\end{array}$

As exponential $f^n$, cannot be negative, so $2^x=y=5$.

Therefore, $x={\log }_{2}5$

Hence, option A is correct.