how to use "In"?
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?
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) ?
\[4132.79\times (1+\frac{.048}{4})^{4t}=8,000\] \[4132.79\times (1.012)^{4t}=8,000\] \[1.012^{4t}=8,000\div 4132.79\]
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?)
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 \[b^x=A\iff x=\frac{\ln(A)}{\ln(b)}\]
in my calculator i have log2, log10, and In... can use any true? (I used In)
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
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
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...
Was 13 years and 10 months the correct answer?
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