Drawing a blank, need help on a math problem

[Whorax]

Question:

You deposit $1200 into an account with 3% annual compound interest. In how many years will the account be growing by $190?

 

I thought the formula I needed to use for this was 190=1200(1.03)^t (ln1.03), however whenever I finish calculating this shit I end up getting t=~181.2. The answer is ~56.7, and I can't for the life of me get an answer that isn't 181.2.

 

Help...please...this is what I get for being on LTT instead of taking notes.

[Speedyv]

Give it a shot solving with $1390

[HeyKitty]

every year a little more is added, so year 1 will add ____ and then year two will add ____ plus extra %3 for what was added last year. To sick to do the math... But I believe that will get you the right answer.

[Bonesy]

OUCH! Been too long since i've been in school...but I can already tell you that for a growth of $190 at a 3% annual compound, you are looking JUST short of 5 years(4.9725 years to be exact)....no idea where the other values are being derived from.

 

Please someone correct me if i'm mistaken, it's been a while, BUT it should be:

 

1200(1+.03/1)1*4.9725=1390

Principal(1+Rate/Interval)^Interval*Length

[Speedyv]

OUCH! Been too long since i've been in school...but I can already tell you that for a growth of $190 at a 3% annual compound, you are looking JUST short of 5 years(4.9725 years to be exact)....no idea where the other values are being derived from.

 

Please someone correct me if i'm mistaken, it's been a while, BUT it should be:

 

1200(1+.03/1)1*4.9725=1390

Principal(1+Rate/Interval)^Interval*Length

 

The values are in months....

 

It comes to be pretty close but not exact to the given answer.

[Bonesy]

The values are in months....

 

It comes to be pretty close but not exact to the given answer.

 

"3% annual compound interest"   technically n(Interval) is not needed for annual compounding, but it was habit for me to always use it anyways, for monthly/quarterly compounds.

[Speedyv]

"3% annual compound interest"

 

Doesn't mean the answer can't be converted to months...

It's the only logical explanation. 

[Bonesy]

Only going off the information OP provided...he states an annual compound, and the question is phrased: In how many YEARS will the account be growing by $190?  Given that information, a 3% annual compound would take 4.9725 years to mature $190

[Whorax]

Doesn't mean the answer can't be converted to months...

It's the only logical explanation. 

Only going off the information OP provided...he states an annual compound, and the question is phrased: In how many YEARS will the account be growing by $190?  Given that information, a 3% annual compound would take 4.9725 years to mature $190

Well guys, I'm an idiot.

 

Here's what I was doing before:

190 = 1200(1.03)^t ln1.03

190/(1200(ln1.03)) = t ln1.03

(190/(1200(ln1.03)))/ln1.03 = t

 

 

What I needed to do was :

190 = 1200(1.03)^t ln1.03

190/(1200(ln1.03)) = t ln1.03

ln(190/(1200(ln1.03)))/ln1.03 = t

56.779 = t

 

I completely forgot to take the ln of the opposite side when I was bringing down the exponent...lol...if that made sense.

[marioo1182]

Well guys, I'm an idiot.

 

Here's what I was doing before:

190 = 1200(1.03)^t ln1.03

190/(1200(ln1.03)) = t ln1.03

(190/(1200(ln1.03)))/ln1.03 = t

 

 

What I needed to do was :

190 = 1200(1.03)^t ln1.03

190/(1200(ln1.03)) = t ln1.03

ln(190/(1200(ln1.03)))/ln1.03 = t

56.779 = t

 

I completely forgot to take the ln of the opposite side when I was bringing down the exponent...lol...if that made sense.

Just worked it out. You got it.

This is grade 12 logarithms isn't it?

[Whorax]

Just worked it out. You got it.

This is grade 12 logarithms isn't it?

Pretty sure, although I'm taking this as..dummy class (lol?) for my final set of math credits in university, since I'm not majoring in a STEM field or anything that involves heavy math.