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

1000(1 + 0.06 / 12)^12t = 1200(1 + 0.04)^t
1.005^12t = 1.2(1.04)^t
12t log(1.005) = log(1.2) + t log(1.04)

That is my steps so far and I think I need to factor out a t but not sure

Answer 2

@Hero @SolomonZelman @jim_thompson5910 @tkhunny

Answer 3

@Loser66

Answer 4

Have you considered subtracting [t log(1.04)] from each side?

Gather the variables to one side!

Answer 5

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

Answer 6

@tkhunny, and then factor out a t?

Answer 7

That's right. Algebra I!!

Answer 8

You can change log(1.005) - log(1.04) = log(1.005/1.04) if you like.

Answer 9

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

Answer 10

Whoops! I missed the 12. Don't do that.

\(t(12\log(1.005) - \log(1.04)) = \log(1.2)\)

It's time to resort to the numerical solution.

Answer 11

OK, so let me write it down the correct way

Answer 12

So what I have is 12t log(1.005) - log(1.004) = log(1.2)
t(12 log(1.005) - log(1.004) = log(1.2)

Answer 13

Then you divide both sides by log(1.005/1.04)

Answer 14

@tkhunny

Answer 15

No, you didn't undo the error.

\(t = \dfrac{\log(1.2)}{12\log(1.005) - \log(1.04)}\)

Answer 16

I'm not sure I follow

Answer 17

If I'm subtracting two logs, then I divide them

Answer 18

t = 12 log(1.005)/log(1.004) = log 1.2

Answer 19

I agree that's the idea, but that is not what we have.

log(a) - log(b) = log(a/b)

12log(a) - log(b) = log(a^12) - log(b) = log((a^12)/b)

Quite a different animal.

Change \(12\log(1.005) = \log\left(1.005^{12}\right) = \log(1.0616778)\)

Answer 20

log(1.0616778) - log(1.004) = log(1.2)

Answer 21

t = log(1.2) / log(1.0616788/log(1.04)

Answer 22

Where did the "t" go? \(t = \dfrac{\log(1.2)}{\log(1.0616788)-log(1.04)} = \dfrac{log(1.2)}{\log\left(\dfrac{1.0616788}{1.04}\right)}\)

Pay VERY close attention to what you are dividing and what you are subtracting.

Answer 23

t is approximately 8.84 years

Answer 24

Do you believe it? It takes about 9 years for 6.17% to catch up to 4% with a 20% lead? Seems okay.

Answer 25

That seems alright.

Answer 26

Well, there you go. Move on to the next one.

Answer 27

Thanks, I posted it.