dx ⇒‌log‌f′(x)‌‌=log‌f(x)+log‌c ⇒‌f′(x)‌‌=cf(x) Now, put x=0, we get ‌f′(0)‌‌=cf(0) ⇒‌2‌‌=c×1 ⇒‌c‌‌=2 Putting the value of c=2 in Eq. (i), we get ‌log‌f′(x)‌‌=log‌f(x)+log‌2 ⇒‌f′(x)‌‌=2f(x)⇒∫‌
f′(x)
f(x)
dx=∫2dx ⇒‌log‌f(x)‌‌=2x+D⇒f(x)=e2x+D ⇒‌f(x)‌‌=eD⋅e2x ⇒‌f(x)‌‌=k⋅e2x [Let k=eD ] Put x=0, we get ‌f(0)‌‌=k⋅e0 ⇒‌1‌‌=k⇒f(x)=k⋅e2x ‌∴‌f(x)‌‌=e2x Put x=1, we get f(1)=e2 Clearly, e2 lies in (6,9).