Question

Can anyone tell me what this is?

Answer 1

https://media.studyisland.com/cgi-bin/mimetex.cgi? \begin{eqnarray}F\%20&=&\%20F_{0}(1\%20+\%20r)^{t}\\\vspace*{9\hspace*{-10}}\\6,643\%20&=&\%204,125(1\%20+\%200.10)^{t}\\\vspace*{9\hspace*{-10}}\\1.61042\%20&\approx&\%20(1.10)^{t}\\\vspace*{9\hspace*{-10}}\\\text{ln}(1.61042)\%20&\approx&\%20\text{ln}[(1.10)^{t}]\\\vspace*{9\hspace*{-10}}\\\text{ln}(1.61042)\%20&\approx&\%20t\,\text{ln}(1.10)\\\vspace*{9\hspace*{-10}}\\\frac{\text{ln}(1.61042)}{\text{ln}(1.10)}\%20&\approx&\%20t\\\vspace*{9\hspace*{-10}}\\5\%20&\approx&\%20t\\\vspace*{9\hspace*{-10}}\\\end{eqnarray}

Answer 2

They didn't teach jack on whatever ln is

Answer 3

I know how to get to 1.61042 but after that everything goes wrong

Answer 4

Ln is a natural log.
\[\ln=\log_e\]for example \[\ln(4)=\log_e4\]

ln is just " log base e"

Answer 5

Ok but how in the world do they get to 5 all the way at the bottom?

Answer 6

idk what the problem is, b/c when I click your watchmen it says "no query string"

Answer 7

your attachment, not watchmen, sorry _:(

Answer 8

F = F0(1+r)^t

6,643 = 4,125(1 + 0.10)^6

1.61042 = (1.10)^t

(This is where I mess up)

ln(1.61042) = ln[(1.10)]^t

ln(1.61042) = tln(1.10)

ln(1.61042) (over) ln(1.10) = t

5 = t

Answer 9

I would suggest t just take log of both sides, not natural log.
\[1.61042 = (1.10)^t\]\[\log~~1.61042~ =\log~~ (1.10)^t~\]\[\log~~1.61042~ =t~~\log~~ (1.10)~\]\[\log~~1.61042~/~~\log~~ (1.10) =t~\]\[\log~~_{1.10}~~~1.61042~ =t~\]

http://www.1728.org/logrithm.htm

Answer 10

ask if you have any questions.