Question

The number of solutions of the equation $n^{-\left|x\right|}\left|m-\left|x\right|\right|=1$ (where$m,n>1$ and $n>m$) is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C

$2$

Explanation for correct option

Given that $n^{-\left|x\right|}\left|m-\left|x\right|\right|=1$

$\begin{array}{crcll}\therefore & n^{-\left|x\right|}\left|m-\left|x\right|\right|& =& 1& \\ \Rightarrow & \log \left(n^{-\left|x\right|}\left|m-\left|x\right|\right|\right)& =& \log \left(1\right)& \\ \Rightarrow & \log \left(n^{-\left|x\right|}\right)+\log \left(\left|m-\left|x\right|\right|\right)& =& 0& \mathbf{}\left[\because log\left(ab\right)=log\left(a\right)+log\left(b\right)\right]\\ \Rightarrow & -\left|x\right|\log \left(n\right)& =& -\log \left(\left|m-\left|x\right|\right|\right)& \mathbf{}\left[\because log\left(a^b\right)=blog\left(a\right)\right]\\ \Rightarrow & \left|x\right|\log \left(n\right)& =& \frac{\log \left(m\right)}{\log \left(\left|x\right|\right)}& \\ \Rightarrow & \frac{\log \left(n\right)}{\log \left(m\right)}& =& \frac{1}{\left|x\right|\log \left(\left|x\right|\right)}& \\ \Rightarrow & \log \left(n-m\right)& =& \frac{1}{\left|x\right|}\log \left(-\left|x\right|\right)& \mathbf{}\left[\because \frac{log\left(a\right)}{log\left(b\right)}=log\left(a-b\right)\right]\\ \Rightarrow & \log \left(n-m\right)& =& \log {\left(-\left|x\right|\right)}^{\frac{1}{\left|x\right|}}& \\ \Rightarrow & \left(n-m\right)& =& {\left(-\left|x\right|\right)}^{\frac{1}{\left|x\right|}}& \end{array}$

Given that $n>m\Rightarrow \left(n-m\right)>0$

For the above condition to be true the following are the possible solutions:

1. $\left|x\right|=-x\Rightarrow x<0$
2. $\frac{1}{\left|x\right|}=2k\Rightarrow \left|x\right|=\frac{1}{2k}$ where $k$ is any integer

Thus the given equation has two solutions.

Hence, the correct option is option(C)