Answer:
t= 11.9
Step-by-step explanation:
Compounded Continuously:
A=Pe^{rt}
A=Pe
rt
A=390\hspace{35px}P=290\hspace{35px}r=0.025
A=390P=290r=0.025
Given values
390=
390=
\,\,290e^{0.025t}
290e
0.025t
Plug in
\frac{390}{290}=
290
390
=
\,\,\frac{290e^{0.025t}}{290}
290
290e
0.025t
Divide by 290
1.3448276=
1.3448276=
\,\,e^{0.025t}
e
0.025t
\ln\left(1.3448276\right)=
ln(1.3448276)=
\,\,\ln\left(e^{0.025t}\right)
ln(e
0.025t
)
Take the natural log of both sides
\ln\left(1.3448276\right)=
ln(1.3448276)=
\,\,0.025t
0.025t
ln cancels the e
\frac{\ln\left(1.3448276\right)}{0.025}=
0.025
ln(1.3448276)
=
\,\,\frac{0.025t}{0.025}
0.025
0.025t
Divide by 0.025
11.8506326=
11.8506326=
\,\,t
t
t\approx
t≈
\,\,11.9
11.9
Round to the nearest tenth of a year
