Question

Please anyone? Medal and Fan! Solving Exponential Equations with Unequal Bases.

Answer 1

You are going to purchase a new car, but being a responsible consumer means doing a little bit of research first.

First, you find the vehicle you are purchasing and its price.
Vehicle: Chevy Volt
Price: $39,145

Answer 2

Current interest rate: 3%

Answer 3

Using the function A(t)=P(1+r/n)^nt, create the function that represents your new car loan that is compounded monthly. The principle will be the price of the vehicle you selected, not how much you are putting down.

Answer 4

I have a function already. A(t) = 39,145(1+ 0.03/12)^(12t).

Answer 5

"principal" not "principle"

Answer 6

Being a smart financial planner, you want to figure out how many months it will be until your principal is paid down to $10,000.00. Solve for t and show all of your work. Note that t will be negative because the number of months will decrease the principal.

Answer 7

I agree with your formula so far

Answer 8

so pretty much you'd solve for "t"..... setting A(t) = 10,000 it seems

Answer 9

Why 10,000?

Answer 10

Oh wait

Answer 11

Lol

Answer 12

blonde moment

Answer 13

the A(t) amount accrued will be less than the principal
thus the "years" will be negative, so seems like you're getting 10,000 to years before you even asked for the loan =)

Answer 14

Okay I'm very confused with the second part. Because I've got my function but I have no idea how to put it to use with the second part.

Answer 15

Okay

Answer 16

This formula doesn't actually take any payments into account, however...

Answer 17

so say, this year is year 1, I asked for a loan

so year -1 is last year, year -2 is two years ago, year -3 is three years ago

Answer 18

the Amount accrued by definition, for an accumulated interest, cannot be less than the principal

however, in this case 10,000 is less than 39,145, thus "t" , years, will end up being negative

Answer 19

Okay I understand why it would be negative now

Answer 20

I'm not fully understanding how to solve for t. I've been trying for an hour now but my brain is just going "nope not today"

Answer 21

let's try your formula with say the 1st year, t = 1 \(\bf A(t)=39,145\left(1+\frac{0.03}{12}\right)^{12t}\implies A(1)=39,145\left(1+\frac{0.03}{12}\right)^{12}
\\ \quad \\
\implies A(1)\approx 40335.6326\)

so you see, the A(t) is at the very 1st year a big amount, way over 10,000

so if we set A(t) to a smaller number than the principal, the "t" will be negative

Answer 22

using the compound formula by itself
it's just a formula, it doesn't differentiate on what you plug to it

if you use a negative "t", you'd end up with a diminished OUTPUT from the given INPUT

Answer 23

SO I would use a negative number for "t"? right? So How do I know/figure out which number to use?

Answer 24

well... you don't, you'd solve for "t", and you'd get a negative value

one sec

Answer 25

okay take your time. I'm trying to figure this out too, I'm very lost sorry. So I think to solve I would start by solving the part of the equation in the parenthesis right? \[(1+\frac{ 0.03 }{ 12 })\]

Answer 26

1.0025. I don't know if this is what I'm supposed to be doing though.
So \[A(t)=39,145(1.0025)^{12t}\]

Answer 27

\(\bf A(t)=39,145\left(1+\frac{0.03}{12}\right)^{12t}\implies 10,000=39,145\left(1+\frac{0.03}{12}\right)^{12t}
\\ \quad \\
\cfrac{10,000}{39,145}=\left(1+\frac{0.03}{12}\right)^{12t}\implies \cfrac{10,000}{39,145}=1.0025^{12t}
\\ \quad \\
\textit{log cancellation rule of }log_{\color{blue}{ a}}{\color{blue}{ a}}^x=x
\\ \quad \\
log_{1.0025}\left(\frac{10,000}{39,145}\right)=log_{{\color{blue}{ 1.0025}}}({\color{blue}{ 1.0025}}^{12t})\implies log_{1.0025}\left(\frac{10,000}{39,145}\right)=12t
\\ \quad \\
\cfrac{log_{1.0025}\left(\frac{10,000}{39,145}\right)}{12}=t\)

Answer 28

Wow I was doing it wrong, I thought so. okay now I see, setting A(t) to 10,000. The last part, do I need to solve that further?

Answer 29

well, the calculator may not have a log base 1.0025 so you'd use the log change of base rule, to say change it to base 10 to get the log value so \(\bf \textit{change of base rule of }log_ab\implies \cfrac{log_{\color{green}{ c}}b}{log_{\color{green}{ c}}a}
\\ \quad \\

log_{1.0025}\left(\frac{10,000}{39,145}\right)\implies \cfrac{log_{{\color{green}{ 10}}}\left(\frac{10,000}{39,145}\right)}{log_{{\color{green}{ 10}}}(1.0025)}\)

Answer 30

Okay I see. So, once that is figured out, that would be the value of "t", right? And like you said before, it should be negative?

Answer 31

yeap

Answer 32

Okay thank you so much!

Answer 33

yw