errr...anyone good at math?

[ManBearPig]

how would i do this?

13e^8t = 15e^9t

thanks

 

[TuxDave]

1) (a^x)/(a^y) = a^(x-y)
2) a=b^c........ log(base b)[a] = c

 

[So]

on your ti 89

solve(13e^8t = 15e^9t, t)

:evil:

 

[jmcoreymv]

What are you trying to do? t=-inf?

 

[KLin]

42 is the answer. How did I come up with this? I don't know

 

[Whoozyerdaddy]

Originally posted by: Heen05
how would i do this?

13e^8t = 15e^9t

thanks

Looks like l33t speak to me...

 

[Syringer]

Originally posted by: TuxDave
1) (a^x)/(a^y) = a^(x-y)
2) a=b^c........ log(base b)[a] = c

:thumbsup:

Divide both sides by the left side, so on the right side you get..

15e^9t/13e^8t = 15/13 * e^t

And that equals one..

So

15/13 * e^t = 1
e^t = 13/15

So x = ln (13/15)

 

[cirthix]

13e^8t=15e^9t

13/15=e^(9t-8t)

e^t=13/15

t=ln(13/15)

 

[ManBearPig]

so...how do i do this?

the effective annual yield is 6.3 %

What is the nominal annual interest rate?

and also how do i find the effective annual yeild if i know the nominal annual interest rate?

thanks soo much

 

[XZeroII]

errr...anyone good at math?

pfff. Do you really have to ask that here?

 

[RichUK]

Ermm ? I can add stuff on a calculator