SOLUTION: if $350 is deposited in an account paying 2% annual interest, compunded continuously, how long will it take for the account to increase to $1500? A=Pe^rt 1500=350e^.02(t)

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: if $350 is deposited in an account paying 2% annual interest, compunded continuously, how long will it take for the account to increase to $1500? A=Pe^rt 1500=350e^.02(t)       Log On


   

Question 193252: if $350 is deposited in an account paying 2% annual interest, compunded continuously, how long will it take for the account to increase to $1500?

A=Pe^rt
1500=350e^.02(t)
then im lost...
Found 2 solutions by MathGuyJoe, jim_thompson5910:
Answer by MathGuyJoe(20) About Me  (Show Source):
You can put this solution on YOUR website!
To Solve: 1500=350e%5E%28.02t%29
You must first isolate the e^x component -- so divide both sides by 350:
+1500+%2F+350+=+e%5E%28.02t%29+
Simplifying gives you:
+30+%2F+7+=+e%5E%28.02t%29+
Next, take the natural log (ln) of both sides:
+ln%2830%2F7%29+=+ln%28e%5E%28.02t%29%29+
Now, use the power rule +ln%28x%29%5En+=+n+%2A+ln%28x%29+ to simplify the right side:
+ln%2830%2F7%29+=+.02t%2Aln%28e%29+
Since +ln%28e%29+=+1+, you're left with:
+ln%2830%2F7%29+=+.02t+
Divide both sides by .02 to get:
+%28ln%2830%2F7%29%2F.02%29+=+t+
So t = 72.76 years


Hope this helps.


~ Joe

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
A=Pe%5E%28rt%29 Start with the given equation.


1500=350e%5E%280.02t%29 Plug in A=1500, P=350, and r=0.02


1500%2F350=e%5E%280.02t%29 Divide both sides by 350.


4.2857=e%5E%280.02t%29 Divide


ln%284.2857%29=0.02t Take the natural log of both sides (to eliminate the base "e")


1.4553=0.02t Evaluate the natural log of 4.2857


1.4553%2F0.02=t Divide both sides by 0.02


72.765=t Divide


t=72.765 Rearrange the equation


So it will take approximately 72.765 years for $350 to grow into $1,500.