i'v never used in my math class before, so to me it doesn't make much sense...

(also i tried it out, and the answer doesn't work out for me...)

i don't think that will work for what i'm trying to do.

since when i plug

Results 1 to 7 of 7

- June 3rd 2007, 08:03 PM #1
## a(b)^x = a(b)^x ( getting x value )

i have a question here and i'm not sure how to tackle this sucker...

Same deposits $800 into an investment fund that erns 8% per year, compounded annually. Jessica deposits $1000 into an investment fund that earns 6% per year, compounded annually. When will their investments be equal( 3 places after the decimal)?

thats the question and heres what i wrote down...

Sam

a= 800

b= 8% = 1.08

c= (null)

x= (missing value)

Jessica

a= 1000

b= 6% = 1.06

c= (null)

x= (missing value)

so i put this:

800(1.8)^x=1000(1.06)^x

and i don't know where to go from there...

*not that this is in the functions and relations section, and we are using logs (not sure if that is needed for this question...)

- June 3rd 2007, 08:16 PM #2

- June 3rd 2007, 08:22 PM #3

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- June 3rd 2007, 08:25 PM #4

- June 3rd 2007, 08:29 PM #5
you actually ended up using the same formula without trealizing it. now you are correct so far. now all that remains is to find . we need logs for this. and you should remember, whenever you see a variable as a power and you need to solve for the variable, chances are you need logs.

let's pick up where you left off.

.....divide both sides by 1000

......take log base 10 of both sides

...since

......group 's on one side

......since

......since

And now just plug that into your calculator

EDIT: Ah, CaptainBlack beat me to it, oh well

- June 3rd 2007, 08:31 PM #6

- June 3rd 2007, 08:33 PM #7