I need help solving the following:

c(1+g1)^n = k(1+g2)^n

where c and k are positive constants and g1 and g2 are growth rates. need to solve for n

c>k g2>g1

I should know how to do this but I am having trouble. Thank you

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- Feb 8th 2011, 05:55 AMalexpollemaUsing Log to solve for variable in the exponent
I need help solving the following:

c(1+g1)^n = k(1+g2)^n

where c and k are positive constants and g1 and g2 are growth rates. need to solve for n

c>k g2>g1

I should know how to do this but I am having trouble. Thank you - Feb 8th 2011, 06:10 AMAckbeet
I would try this:

$\displaystyle \dfrac{c}{k}=\dfrac{(1+g_{2})^{n}}{(1+g_{1})^{n}}= \left(\dfrac{1+g_{2}}{1+g_{1}}\right)^{\!\!n}.$

Can you continue? - Feb 8th 2011, 06:26 AMalexpollema
is this correct then?

[log(c/k)]/log[(1+g2)/1+g1)] =

I appreciate the help. Can you refer me to a source that gives basic log rules? I should have learned them in high school math class but I must not have been paying attention during those classes - Feb 8th 2011, 06:50 AMAckbeet
See here for a list of the basic logarithmic identities.

You don't have an equation in post # 3. So, I would not agree with your result as is. - Feb 8th 2011, 06:53 AMalexpollema
ok I am sorry,

[log(c/k)]/log[(1+g2)/1+g1)] = n - Feb 8th 2011, 06:55 AMAckbeet
You just need one more parenthesis in there. Put it like this:

[log(c/k)]/log[(1+g2)/(1+g1)] = n. Then you have it.