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- Oct 8th 2012, 08:56 PMEraser147Logarithms
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- Oct 8th 2012, 09:10 PMMarkFLRe: Logarithms
What do you think is the best first step?

- Oct 8th 2012, 09:10 PMchiroRe: Logarithms
Hey Eraser147.

Try collecting the 10^(t) terms to one side and then use the relationship that log_10(10^t) = t where log_10(x) = log(x)/log(10) where log(x) is the natural logarithm. - Oct 8th 2012, 09:46 PMEraser147Re: Logarithms
I ended up with $\displaystyle 10^t+1=.8*10^t+1.6$. I don't know where to go from there.

- Oct 8th 2012, 09:53 PMchiroRe: Logarithms
10^(t) - 0.8*10^(t) = 1.6 - 1 which means

0.2*10^(t) = 0.6 which means

10^(t) = 0.6/0.2 = 3. - Oct 8th 2012, 10:12 PMEraser147Re: Logarithms
Oh... THANKS A BUNCH. My error was not realizing I should subtract the 0.8*10^(t)

- Oct 9th 2012, 04:09 AMPlatoRe: Logarithms