Given equation is [ex]2+[ex+1]−3=0
We know that [x+I]=[x]+I,I∈Z, where [⋅], represents the greatest integer function.
⇒[ex]2+[ex]+1−3=0
Let, [ex]=t
⇒t2+t−2=0
⇒(t+2)(t−1)=0
⇒t=−2,1
But [ex]=−2, (Not possible because we know that ex>0)
Hence, [ex]=1 and also, we know that if [x]=1,⇒1≤x<2
∴1≤ex<2
And, now by using the definition of logarithm, we have, if ex=a, then logea=x
⇒loge1≤x<loge2
⇒0≤x<loge2
⇒x∈[0,loge2).