Question

I need 3 & 4 checked to see if I am correct, and help explaining 5 & 6

Answer 1

@amoodarya @jdoe0001 @ehuman @skullpatrol

Answer 2

am I wrong or they seem to be done already

Answer 3

3 & 4 are done already but 5 & 6 aren't. I wanted to make sure 3 & 4 were correct and 5&6 I need help on.

Answer 4

3 and 4 are correct.

Answer 5

yeap, they look fine to me

Answer 6

For 5, replace A(t) with 100,000 and since you are trying to find the time, use t in the equation and solve for t

Answer 7

For 6, you divided by 12 to find the monthly rate. Since there are 52 weeks in a year, I would divide by 52 to find the weekly interest rate.

Answer 8

\(\bf \large {A=p\left(1+\frac{4}{n}\right)^{nt}\quad \textit{solving for "t"}\\ \quad \\
\cfrac{A}{p}=\left(1+\frac{4}{n}\right)^{nt}\\ \quad \\
\textit{log cancellation rule of }a^{log_ax}=x\quad thus\\ \quad \\
log_{1+\frac{4}{n}}\left[\cfrac{A}{p}\right]=log_{1+\frac{4}{n}}\left[\left(1+\frac{4}{n}\right)^{nt}\right]\implies log_{1+\frac{4}{n}}\left[\cfrac{A}{p}\right]=nt\\ \quad \\
\cfrac{log_{1+\frac{4}{n}}\left[\cfrac{A}{p}\right]}{n}=t}\)

Answer 9

I knew it had to do with log! And I'm not understand those... argh! LOL

Answer 10

hmmm.... wait. a sec... lemme recheck... myself =)

Answer 11

I think I even used the wrong cancellation rule =)

Answer 12

\[100000=56000(1+\frac{.0199}{12})^{12t}\]

Answer 13

ohh.. nope, is correct, I just wrote the wrong cancellation rule =), anyhow the log cancellation rule is \(\large \bf log_aa^x=x\)

Answer 14

\[1.7857=1.0017^{12t}\]

Answer 15

\[\log_{10}1.7857=12t(\log_{10}1.0017) \]

Answer 16

\[t=\frac{\log_{10}1.7857}{12(\log_{10}1.0017)} \]

Answer 17

\[100000 = 56230(1+\frac{ .0199 }{ 12 })^{12t}\]\[100000=56230(1+.001658333)^{12t}\]\[100000=56230(1.001658333)^{12t}\]

Answer 18

That was how far I got

Answer 19

Keep going.

Answer 20

Divide both sides by 56230

Answer 21

\[1.77841010 = (1.001658333)^{12t}\]

Answer 22

Take the log of both sides.

Answer 23

I'm getting there.. lol

Answer 24

\[\log_{10} 1.77841010 = 12t(\log_{10} 1.001658333)\]

Answer 25

yep

Answer 26

\[t=\frac{ \log_{10}1.77841010 }{ 12\log_{10}1.001658333 }\]

Answer 27

yep

Answer 28

\[t=\frac{ .2500319162 }{12(7.196083607E-4)}\]
That's what it's saying when I put the log10 in for each of them
What is the E-4

Answer 29

It means move the decimal point 4 place to the left. It is giving you the number in scientific notation

Answer 30

So...\[t=\frac{ .2500319162 }{ .00863530032 }\]

Answer 31

t = 28.9546

Answer 32

So it would be 28.95 years to owe $100,000?

Answer 33

That's what I got. I would probably round it to 29 years.

Answer 34

Ok... now that I understand #5, on to #6... are they just asking what the interest rate would be on a weekly basis? Which would be 1.99%/52

Answer 35

yes

Answer 36

Thank you so much @Mertsj !

Answer 37

yw