Math Help - logarithmic word problem

1. logarithmic word problem

Hello, here is my problem:

Alex invests $50,000 at an interest rate of 0.7% compounded monthly. Laura invests$40,000 at 9.5% compounded annually. After how many years will the two investments be equal in value?

okay so this is what I have so far

50,000 x (.07x12)^n=x

40,000 x .095^n=x

I am not really sure how to set up the equation and I'm pretty sure I started off wrong, help is much appreciated!

Thank you!
Dan

2. Re: logarithmic word problem

The 1st amount gets multiplied by 1.007 each month and the 2nd amount by 1.095 each year.

3. Re: logarithmic word problem

it should have been: 50,000 x (1.07^n)12=x and 40,000 x (1.095 ^n)=x I think im figuring it out. Biffboy, i think you ment to write 1.07

4. Re: logarithmic word problem

If the interest rate is 0.7% we multiply by 1.007. If you meant 7% we multiply by 1.07. In your question you said 0.7%
Also n years =12n months so with the 50000 the index will be 12n

5. Re: logarithmic word problem

The reason i cant figure it out is because it has two variables. In the answers n= 10.6, but if you sub that in the x variable in each equation are different when they should be the same according to the question in the first post. I am clearly missing something...

6. Re: logarithmic word problem

The first rate can't be 7% per month as amounts then would never be equal. So it must be 0.7% and you should be multiplying by 1.007^(12n)

7. Re: logarithmic word problem

I made a mistake writing the problem, sorry it is 7% not 0.7%, and 9.5%. My bad... Thank you for you help though, I'm busting out the calculator right now!

8. Re: logarithmic word problem

Originally Posted by Dwahl
Hello, here is my problem:

Alex invests $50,000 at an interest rate of 0.7% compounded monthly. Laura invests$40,000 at 9.5% compounded annually. After how many years will the two investments be equal in value?

okay so this is what I have so far

50,000 x (.07x12)^n=x
No "7% compounded monthly" does NOT mean 7% per month. It is still 7% per year which is $.07/12= 0.00583$ percent per month. In n years, there are, of course, 12n months so it will be $50000(1.00583)^{12n}$

40,000 x .095^n=x
And this should have been $40000(1.095)^n$
Set them equal and solve for n.

I am not really sure how to set up the equation and I'm pretty sure I started off wrong, help is much appreciated!

Thank you!
Dan

9. Re: logarithmic word problem

this is what i have so far, thanks for all your help guys, I really appreciate it!

50000 (1.00583)^12n = 40000 (1.095)^n

log(50000 x 1.00583^12n) = log(40000 x1.095^n)

log 50000 + log 1.00583^12n = log 40000 + log 1.095^n

log 50000 + 12n log 1.00583= log 40000+ n log 1.095

and this is sort of where i get a bit lost, below is what i did.

log 50000 - log 40000= -12n log 1.00583 + n log 1.095

Im not sure how to simplify it anymore. so i tried this

__ log 50000- log 40000____ = n+n
-12 + log 1.00583 + log 1.095

Can I get a little more help simplifying it, because I know I did it wrong :/ thanks again!

10. Re: logarithmic word problem

Last line wrong. In the previous line write the right hand side as n(log1.095-12log1.00583)