Let [x] denote the greatest integer less than or equal to x.

Question

Let $[x]$ denote the greatest integer less than or equal to $x$. Then the values of $x\in\mathbb R$ satisfying $[e^x]^2+[e^x+1]-3=0$ lie in the interval

Accepted Answer

$[e^x]^2+[e^x]+1-3=0 \implies [e^x]^2+[e^x]-2=0 \implies ([e^x]-1)([e^x]+2)=0$. Since $e^x>0$, $[e^x]
e-2$, so $[e^x]=1 \implies 1\le e^x<2 \implies 0\le x<\log_e2.$

Solutions · 1