Step-by-step explanation:
Given
No. of employees [tex]N(t)=500(0.02)^{0.7^{t}}[/tex]
(a)When the company opens i.e. at t=0
[tex]N(0)=500(0.02)^{1}=10[/tex]
There are 10 employees at the beginning
(b)when N=100
[tex]100=500(0.02)^{0.7^{t}}\\0.2=0.02^{0.7^{t}}[/tex]
Taking log both sides
[tex]\ln (0.2)=0.7^t \cdot \ln (0.02)[/tex]
[tex]0.7^t=\frac{\ln (0.2)}{\ln (0.02)}=0.4114[/tex]
again taking log
[tex]t\times \ln (0.7)=\ln (0.4114)[/tex]
[tex]t=\dfrac{\ln (0.4114)}{\ln (0.7)}=2.49[/tex]
(c)when t tends to [tex]\infty[/tex], [tex](0.7)^t\rightarrow 0[/tex]
So, [tex]N(\infty)=500\cdot(0.02)^0=500\times1=500[/tex]
i.e. there can be 500 employees at max
