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8000 = 4132.79(1+.048/12)^12*t
8000 = 4132.79(1.004)^12t (now I get stuck here..

\[4132.79\times (1+\frac{.048}{4})^{4t}=8,000\]

says QUARTERLY so be careful
usually it is monthly

yah u right... misread

but then how do you continue... because that's when I get lost :/

divide by 4132.78

then use the change of base formula
\[b^x=A\iff x=\frac{\ln(A)}{\ln(b)}\]

what actually In use for... like, I understand nothing at all about In... (In=log right) ?

and why is it use in this kind of formulas?

\[1.012^{4t}=1.936\] rounded

the to solve for \(4t\) use
\[4t=\frac{\log(1.936)}{\log(1.012)}\]

it doesn't matter what log you use

mmm... but why log? like.. what is its function? I never understood when the teacher explained it...

then i am sure i cannot explain it in a chat box here but basically
\[b^x=y\iff \log_b(y)=x\]

thats the formula right?

it is a way to solve for a variable that is in the exponent

mmmm.. got it.. the answer I got for the formula is t= 10.8 (then years and 8 month?)

in my calculator i have log2, log10, and In... can use any true? (I used In)

haha got it...
so my answer is 10.8... how do i convert that into years and months?

sorry ... i got 13.8

makes no difference, all logs are the same
\[\log_b(x)=\frac{\log_a(x)}{\log_a(b)}\]

got it...
but how do you convert t = 13.85 into years and months?

what does \(t\) represent?

time...

so im guessing it would be 13 years and 8 months right?

in what units?

years&months

yes t is time in years

Was 13 years and 10 months the correct answer?