SOLUTION: If interest is compounded continuously at the rate of 3% per year, approximate the number of years it will take an initial deposit of $7000 to grow to $27,000. (Round your answer t

Click here to see ALL problems on logarithm

Question 1149204: If interest is compounded continuously at the rate of 3% per year, approximate the number of years it will take an initial deposit of $7000 to grow to $27,000. (Round your answer to one decimal place.)

Found 2 solutions by addingup, ikleyn:
Answer by addingup(3677)   (Show Source):
You can put this solution on YOUR website!
t = ln(A/P)/ln(1 + r) = [ln(A) - ln(P)]/ln(1 + r)
t = ln(27,000/7,000)/ln(1+0.03) = [ln(27,000 - ln(7,000]/ln(1+0.03)
t = [ln(27,000 - ln(7,000]/ln(1+0.03)
t = (10.2 - 8.85)/0.02956
t = 45.67
--------------------------------------
Check:
7,000(1.03)^45.67 = 27,000.64 round to a whole number: 27,000
.
Happy learning

Answer by ikleyn(53853)   (Show Source):
You can put this solution on YOUR website!
.
If interest is compounded continuously at the rate of 3% per year, approximate the number of years
it will take an initial deposit of $7000 to grow to $27,000. (Round your answer to one decimal place.)
~~~~~~~~~~~~~~~~~~~~~~~~~

        The solution and the answer in the post by @addingup to both problems are incorrect.
        It is because his starting setup equation is incorrect, which is a conceptual/strategic error.

        I came to bring a correct solution.

The setup equation for this continuously compounded account is 27000 = , <<<---=== It is a standard translation for a continuously compound account where 't' is the time in years. Divide both sides by 7000 to get = . Take natural logarithm of both sides = 0.03*t. Express 't' and calculate t = = 44.9976 years. It is reasonable to round it to the closest integer, which is 45 years. ANSWER. 45 years.

Solved correctly, so you can learn/teach safely from my post.

Ignore the post by @addingup, since it is wrong solution and wrong teaching.