Tech Now, a technology company, began in 1980 with 250 customers. Since 1980, the number of customers is increasing. The following function represents the number of Tech Now customers t years after 1980, where r is the rate of growth. If the company has 1,959 customers 11 years after 1980, then what is the approximate rate of growth?
+118613 y=Ae^(kt)
when t=0 y= 250 SO A=250
y=250e^(kt)
When t=11 y= 1959
$$\\1959=250e^{11k}\\\\
(1959/250)=e^{11k}\\\\
ln(1959/250)=11k\\\\
k=\frac{ln(1959/250)}{11}\\\\$$
$${\frac{{ln}{\left({\frac{{\mathtt{1\,959}}}{{\mathtt{250}}}}\right)}}{{\mathtt{11}}}} = {\mathtt{0.187\: \!157\: \!136\: \!371\: \!117\: \!7}}$$
growth rate is approx 18.7% per annum
Thanks for checking Chris :)
+118613 y=Ae^(kt)
when t=0 y= 250 SO A=250
y=250e^(kt)
When t=11 y= 1959
$$\\1959=250e^{11k}\\\\
(1959/250)=e^{11k}\\\\
ln(1959/250)=11k\\\\
k=\frac{ln(1959/250)}{11}\\\\$$
$${\frac{{ln}{\left({\frac{{\mathtt{1\,959}}}{{\mathtt{250}}}}\right)}}{{\mathtt{11}}}} = {\mathtt{0.187\: \!157\: \!136\: \!371\: \!117\: \!7}}$$
growth rate is approx 18.7% per annum
Thanks for checking Chris :)
