Question

One savings account begins with $1000 and earns 6% interest, compounded monthly. Another account begins at the same time with $1200 and earns 4% interest, compounded yearly. How long will it take for the two accounts to reach the same value? Give an answer in years, rounded to the nearest whole number.

Answer 1

How do I determine when the two accounts reach the same value?

Answer 2

A1 = 1000(1+0.06/12)^12t
A2 = 1200(1+0.04)^t

set A1=A2,
take logarithms and solve for t, the time in years

Answer 3

before you take logs of A1 and A2, do you first divide?
@douglaswinslowcooper

Answer 4

I am stuck at the step where you take logs of the two accounts

Answer 5

Either way should work.
Dividing by 1000 simplifies a bit.

(1.005)^12t = 1.2(1.04)^t

12 t log (1.005) = log(1.2) + t log(1.04)

t [12 log(1.005)-log(1.04)] = log 1.2

t[0.026 -0.017] = 0.079
t = 8.8 years = 9 years

check
1000(1.005)^108 = 1714
1200(1.04)^9 = 1708

close enough for our purposes

Answer 6

Glad to have been of help.

Answer 7

@douglaswinslowcooper
Can you explain the first step when you simplify?

Answer 8

(1.005)^12t = 1.2(1.04)^t
I see where you log both sides

Answer 9

about tree fiddy

Answer 10

1000/1000=1 and 1200/1000=1.2 if that is the step you mean.
I divided both sides by 1000.

Answer 11

the second one you would want to divide by 1200... right?

Answer 12

@douglaswinslowcooper

Answer 13

dividing both sides of the equation by 1000 just to make the numbers shorter