2x = 400 9
Start with \[\large A=Pe^{rt}\]
Not quite. There was an error in your solving.
2=e^rt ln 2 = rt
How did you get that?
I don't think simple interest applies here. That wouldn't require logarithms. 7.2 percent is close to what I got. Did you round off anywhere along the way?
7.2 is a good approximation using the rule of 72, but it is less precise than using the exponential formula.
Not using logarithms. How did you get the 7.2? Did you use the t=72/r approximation?
Ok, that's good, that gives a good estimate, but to be more precise, use the exponential. \[\large A=Pe^{rt}\] A=8000, P=4000, t=10 A/P=2 \[\large \rightarrow 2=e^{10r}\]
Taking the natural log of both sides: \[\large ln(2) = 10r \rightarrow r=\frac{ln(2)}{10}\]
No, P=4000
No, you need to solve 2=e^rt, which has been solved to get r=ln(2)/10. You need to evaluate ln(2)/10 to get a value for r.
That's better. (or even 6.93%) See how that is close to the 7.2% estimate?
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