# How to isolate "R" (from a log equation)

• Dec 8th 2010, 12:36 PM
TN17
How to isolate "R" (from a log equation)
• Dec 8th 2010, 01:03 PM
pickslides
A pretty good attempt, well done!

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

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

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

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

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

$\displaystyle 100(\sqrt[6]{2.5}-1) =R$
• Dec 8th 2010, 04:26 PM
TN17
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 PM
skeeter
$\displaystyle 25000 = 10000\left(1+\frac{R}{100}\right)^{0.6}$

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

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

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

$\displaystyle 100[(2.5)^{\frac{5}{3}}-1] =R$
• Dec 8th 2010, 06:34 PM
TN17
Thank you.
I know where I went wrong now.