how to use "In"?

- anonymous

how to use "In"?

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- anonymous

I have this question.. and I know i Have to use In... but dont know how
Brooklyn has a goal to save $8,000 to buy a new entertainment system. In order to meet that goal, she deposited $4,132.79 into a savings account. If the account has an interest rate of 4.8% compounded quarterly, approximately when will Brooklyn be able to make the purchase?

- anonymous

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

- anonymous

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

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## More answers

- anonymous

says QUARTERLY so be careful
usually it is monthly

- anonymous

yah u right... misread

- anonymous

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

- anonymous

divide by 4132.78

- anonymous

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

- anonymous

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

- anonymous

\[4132.79\times (1+\frac{.048}{4})^{4t}=8,000\]
\[4132.79\times (1.012)^{4t}=8,000\]
\[1.012^{4t}=8,000\div 4132.79\]

- anonymous

and why is it use in this kind of formulas?

- anonymous

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

- anonymous

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

- anonymous

it doesn't matter what log you use

- anonymous

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

- anonymous

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

- anonymous

thats the formula right?

- anonymous

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

- anonymous

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

- anonymous

but you only have two logs on your calculator, \(\log_{10}(x)\) log base ten
and \[\ln(x)=\log_e(x)\] which is log base e
so if you want an actual decimal for an answer, you have to use
\[b^x=A\iff x=\frac{\ln(A)}{\ln(b)}\]

- anonymous

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

- anonymous

in other words, to solve for the variable in the exponent, it is the log of the total divided by the log of the base
that is how to solve for a variable that is in the sky

- anonymous

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

- anonymous

sorry ... i got 13.8

- anonymous

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

- anonymous

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

- anonymous

what does \(t\) represent?

- anonymous

time...

- anonymous

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

- anonymous

in what units?

- anonymous

years&months

- anonymous

yes t is time in years

- anonymous

but in the choices I've got there is no 13 years and 8 month... (they have 13 years and 10 months) so I'm guessing is that one.. but they also have 13 years and 5 months... so like, i want to know how to solve it correctly...

- anonymous

Was 13 years and 10 months the correct answer?

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