Question 541872
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Ok, that is the correct set up for the 'compounded quarterly' scenario.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(1\ +\ \frac{0.06}{4}\right)^{4t}\ =\ 3]


Do the indicated arithmetic and then take the log of both sides (any base, doesn't matter)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(1.015)^{4t}\ =\ \ln(3)]


Use


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x^n)\ =\ n\log_b(x)]


to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4t\,\cdot\,\ln(1.015)\ =\ \ln(3)]


Then a little straight-up algebra:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{\ln(3)}{4\,\cdot\,\ln(1.015)}]


The monthly one works out the same way except that you start with:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(1\ +\ \frac{0.06}{12}\right)^{12t}\ =\ 3]


And end up with


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{\ln(3)}{12\,\cdot\,\ln(1.005)}]


Both of them are 18+ years, but the more frequent compounding gets you to your goal about 30 days earlier.


The reason that the logarithm base doesn't matter is that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\ln(\alpha)}{\ln(\beta)}\ =\ \frac{\log(\alpha)}{\log(\beta)}\ =\ \frac{\log_b(\alpha)}{\log_b(\beta)}]


for all real *[tex \LARGE \alpha\ >\ 0] and *[tex \LARGE \beta\ >\ 0] 


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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