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- Dec 8th 2010, 12:36 PMTN17How to isolate "R" (from a log equation)
- Dec 8th 2010, 01:03 PMpickslides
A pretty good attempt, well done!

I prefer not to introduce the $\displaystyle \displaystyle \log$

$\displaystyle \displaystyle 25000 = 10000\left(1+\frac{R}{100}\right)^6$

$\displaystyle \displaystyle 2.5 =\left(1+\frac{R}{100}\right)^6$

$\displaystyle \displaystyle \sqrt[6]{2.5} =1+\frac{R}{100}$

$\displaystyle \displaystyle\sqrt[6]{2.5}-1 =\frac{R}{100}$

$\displaystyle \displaystyle 100(\sqrt[6]{2.5}-1) =R$ - Dec 8th 2010, 04:26 PMTN17
Thanks. :)

Ah, okay.

How would I do b), though?

I tried the same method, but I didn't get 360.5%

Would I still use 25 000 and 10000, or is that only for the first 6 years? - Dec 8th 2010, 05:27 PMskeeter
$\displaystyle \displaystyle 25000 = 10000\left(1+\frac{R}{100}\right)^{0.6}$

$\displaystyle \displaystyle 2.5 =\left(1+\frac{R}{100}\right)^{0.6}$

$\displaystyle \displaystyle (2.5)^{\frac{5}{3}} =1+\frac{R}{100}$

$\displaystyle \displaystyle (2.5)^{\frac{5}{3}}-1 =\frac{R}{100}$

$\displaystyle \displaystyle 100[(2.5)^{\frac{5}{3}}-1] =R$ - Dec 8th 2010, 06:34 PMTN17
Thank you.

I know where I went wrong now.