# Find the amount A in an account (to the nearest dollar) after 5 years if (dA/dt)=rA, A(0)=800 and A(10)=1800

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dA/dt = rA

(1/A)dA = rdt

Now lets integrate the both sides

ln(A)-lnC = rt

I have choosen the constant as -lnC for both sides after integration.

Then ln(A/C) = rt

(A/C)= e^(rt)

But A(0) = 800

Then 800/C = e^(r*0) since A(0) means A(t) at t=0.

Therefore C= 800/(e^0) = 800/1 = 800

Then A(10) = 1800

So (A/C) = e^(rt)

Then 1800/800 = e^(r*10)

Taking ln in both sides gives you;

ln(1800/800) = 10r

0.81= 10r

0.081= r

So (A/C) = e^(rt) finally become;

A/800 = e^(0.081*t)

So at 5 years t=5

A/800 = e^(0.081*5)

A = 1199.44

**So you can say after five years we have 1200$ in the account.**

The amount A in an account after 5 years if `(dA/dt)=r*A` , `A(0)=800` and `A(10)=1800` has to be determined.

`(dA)/(dt)=r*A`

=> `(1/A)*dA = r*dr`

integrating both the sides

`int(1/A)*dA = int r*dt`

=> `ln A = r*t + C`

To evaluate the variables r and C use A(0)=800 and A(10)=1800

A(0)=800

=> `ln 800 = r*0+ C`

=> C = ln 800

A(10) = 1800

=> `ln 1800 = r*10 + ln 800`

=> `ln(1800/800) = 10*r`

=> `r = (ln(1800/800))/10`

=> `r ~~ 0.0810`

After 5 years the amount in the account is A with `ln A = 0.081*5 + ln 800`

=> `ln A = 7.09`

=> `A = 1200`

**The amount in the account after 5 years is $1200**