Mathematics - Functions Question with Solution | TestHub

Given equation is ${\left[e^x\right]}^2+\left[e^x+1\right]-3=0$

We know that $\left[x+I\right]=\left[x\right]+I, I\in Z,$ where $\left[·\right],$ represents the greatest integer function.

$\Rightarrow {\left[e^x\right]}^2+\left[e^x\right]+1-3=0$

Let, $\left[e^x\right]=t$

$\Rightarrow t^2+t-2=0$

$\Rightarrow \left(t+2\right)\left(t-1\right)=0$

$\Rightarrow t=-2, 1$

But $\left[e^x\right]=-2,$ $($Not possible because we know that $e^x>0)$

Hence, $\left[e^x\right]=1$ and also, we know that if $\left[x\right]=1, \Rightarrow 1\le x<2$

$\therefore 1\le e^x<2$

And, now by using the definition of logarithm, we have, if $e^x=a,$ then ${\log }_{e}a=x$

$\Rightarrow {\log }_{e}1\le x<{\log }_{e}2$

$\Rightarrow 0\le x<{\log }_{e}2$

$\Rightarrow x\in \left[0, {\log }_{\mathrm{e}}2\right).$