Question

I bought a new guitar for $800. It depreciates in value by 8% a year. Write an equation to model the great value. What will be its value in 5 years?

Answer 1

This is what I got so far. G=800(.08)^n-1

Answer 2

and then G=800(.08)^5-1

Answer 3

So what do you get for an answer? And does it make sense?

Answer 4

the thing is i am not sure if i did this right

Answer 5

Well you need to test your equation to see if it seems logical.

And thanks for sowing your thought process.

Answer 6

showing*

Answer 7

np. right now this is what i remember doing a few day back for it but forgot whether i got it all of whether i am missing something.

Answer 8

So using your thought process, what would the the value of the guitar at the end of ONE year.

Answer 9

4.12=G=800(.08)^2-1

Answer 10

?

Answer 11

Your equation would be...

G = 800(.08)^(1-1)

which would be G= 800(0.08)^(1-1)
= 800(0.08)^0
= 800(1)
= 800
so after one year your equation would say NO depreciation, which can't be correct.

Answer 12

oops sorry

Answer 13

i gave year 2 by accident

Answer 14

The exponent ideal is correct, but there can't be any minus 1. So, just use 1 for one year or 5 for five years. Okay?

Answer 15

ok so G=800(.08)^5-1=64^4=$16,777,216

Answer 16

is that correct

Answer 17

This is an alternative approach:
If the guitar depreciates by 8% per year, then its value after 1 year would be (1.00-0.08)*800.
Equivalently, value(1)=$800(0.92)^1

The value after 2 years would be value(2)=$800(0.92)^2.

What would the value be after 5 years?

value(5)=?

Answer 18

Do the exponent before the multiply by 800.

Answer 19

Yes; that follows "order of operations" rules.

Answer 20

The equation format should look like ...

G=800(some percentage depreciation)^N

Answer 21

yes so did i do it right?

Answer 22

No

Answer 23

i did the formula

Answer 24

G=800(.08)^5-1

Answer 25

Do the exponent calculation before the multiply by 800. Remember PEDMAS.

Answer 26

G=800(some percentage depreciation)^N

Answer 27

then 800(.08)=64^5-1

Answer 28

Hold the $800!! Focus on (0.92)^5 first.

Answer 29

oh

Answer 30

You are attempting to evaluate this 800(0.08)^5 from left to right. DON'T.

You MUST do the exponentiation first. What is (0.92)^5?

Answer 31

(.08)^4=4.096e^-5

Answer 32

4.096e^-5(800)

Answer 33

resulting in ... ?

Please note: the value at the beginning is $800.
The value after 1 year is $800(0.92)^1.
tHE VALUE after 5 years is $800(0.92)^5 (not 4).

Answer 34

this =.032768

Answer 35

Once more, Andrew: Please evaluate (0.92)^5

Answer 36

yes but it is ^5-1

Answer 37

You get 0.32768; I get 0.659.

Answer 38

so that would be ^4

Answer 39

this is why i am confused

Answer 40

why?

Andrew, please go back and read what I've typed. ^4 will give you the value of the guitar after 4 years. Is that what you want?

Are you reading what I'm typing?

Answer 41

i am trying tounderstand show me the work you did please so i see what you me

Answer 42

Does this make sense to you? The value of the guitar at the beginning is $800(0.92)^0.

What is (0.92)^0?

Answer 43

1

Answer 44

Everything I did I have typed in for your benefit. You'll need to re-read some of our conversation. Or, just answer my qeustion, above.
Right. So, after 0 years, the value is $800(0.92)^0 = $800(1) = $800. Right or not?

Answer 45

yes

Answer 46

In other weords, the value right now is $800.
Fine.
What is the value after 1 year? $800(0.92)^1 = ??

Answer 47

736

Answer 48

Great. What is the value after 3 years?

Answer 49

622.9504

Answer 50

Right you are.
What is the value afer 5 years?

Answer 51

527.2652186

Answer 52

Excellent. You're done. You're correct. Congrats.
You do need to round that off to $527.27.

Answer 53

ok

Answer 54

Happy?

Answer 55

so what would be the full equation for it?

Answer 56

just 800(.08)^5?

Answer 57

You can deduce that from the pattern we've used.

The value of the guitar after 4 years is $800(0.92)^4.
The value after x years is $800(0.92)^ ??

Answer 58

even though the formula is 800(.08)^n-1

Answer 59

That formula would be correct only if you use a different interpretation.

Example: The value of the guitar at the beginning of year 1 is $800(0.92)^(1-1).
What is this value?

Answer 60

that is what the book showed me i was just trying to use this this is how i got confused i guess

Answer 61

that's why it's so important to DEFINE YOUR TERMS.

Answer 62

Book uses "at the beginning of year 1": $800(0.92)^(1-1)

I use "at the end of the first year:" $800(0.92)^1

Answer 63

so for the 5 years in this formula would be 800(.08)^(6-1)?

Answer 64

In truth I should have followed your book's explanation, but I do not have a copy.

To answer your question, you must define your formula. Are you talking about "at the beginning of year 1" or "at the end of year one"?

Answer 65

?

Answer 66

If you preface your previous statement with: "At the beginning of year 6, the value is..." and your formula will give you $800(0.92)^5 (which is corr4ect).

If you preface

Answer 67

if you preface your prevsious statement with "At the end of year 5, the value is $800(0.92)^5, you'll get the same (correct) result.

Answer 68

so i am right?

Answer 69

for that type of formula

Answer 70

?

Answer 71

Andrew, please read what I said. You have to explain your terms.

Either you are talking about "the beginning of year x)"

or "the end of year 5". Take your pick.

The formulas are almost, but not quite, the same.

Again I'm sorry I didn't have your book available, so I would have followed the book example instead of my own. But both formulas are correct IF AND ONLY IF you define your terms as explained above.

Answer 72

can you try to use the formula i used from the book G=800(.08)^n-1

Answer 73

if i do not use it i am going to get it wrong

Answer 74

Yes.
First, I define my perspective.

"At the BEGINNING of year 1, the value of the guitar is \[$800(0.92)^{1-1}.\]

Answer 75

ok

Answer 76

At the beginning of year 3, the value is \[$800(0.92)^{3-1}, or. $800(0.92)^2\]

Answer 77

yes

Answer 78

so for year 5

Answer 79

So after 5 years, we are at the beginning of year 6, right?

would you please write the correct expression for the value then?

Answer 80

G=800(.08)^(6-1)

Answer 81

?

Answer 82

Your formula uses the "at the beginning of year #" definition.

Now imagine, we're at the end of year 5, which is the beginning of year 6.

So, you MUST use 6 for n.

What is the value of the guitar after 5 years?

Answer 83

Regarding your "G=800(.08)^(6-1)" : What happened to our 0.92?

Answer 84

it has been .08 the whole time

Answer 85

8% not 92%

Answer 86

Not so. Please go back and review our discussion.

The guitar depreciates at the rate of 8% per year. True!! But if you want to know the value of the guitar in the beginning of the nth year, you write $800(1-0.08)^(n-1), or $800(0.92)^(n-1).

Answer 87

?

Answer 88

Depreciates = loses value
Depreciation rate = 0.08 per year
Factor with which to find Value AFTER depreciation = (1-0.08) = 0.92

Answer 89

You calculated the value after 1 year. What was it? go back and find this value.

Answer 90

idk i am too confused

Answer 91

You found that it was $736 and you were right.

Now I'd like for y ou to calculate $800(0.08)^1.


Next, I'd like for you to calculate $800(1-0.08)^1, or $800(0.92)^1.

please go ahead.

Which one produces the correct answer, $736?

Answer 92

the bottom two equations

Answer 93

OK. One more thing. Calculate $800(0.08)^1 again.

Answer 94

64

Answer 95

Yes. That, Andrew, is the DEPRECIATION.

wHAT IS $800-$64?

Answer 96

800-64=736

Answer 97

That's right. That's exactly what you got earlier. The avlue after 1 year is $736. That's found by taking the initial value, $800, and subtracting the depreciation, $64.

Does this make things any clearer?

Neither $800(0.08)^n nor $800(0.08)^(n-1) will give y ou the value of your guitar.

But either $800(0.92)^(whatever) will.

Answer 98

Where did that 0.92 come from? Answer: we subtract the annual depreciation rate (8%) from 100%. or subtract 0.08 from 1.00. Result: 0.92.