Question

The Mendes family bought a new house 11 years ago for $100,000. The house is now worth $196,000. Assuming a steady rate of growth, what was the yearly rate of appreciation? Round your answer to the nearest tenth of a percent (1.2% etc)

Answer 1

@saseal

Answer 2

growth : y = a(1+r)^2

Answer 3

yea

Answer 4

\[196000 = 100000 (1 + \frac{ r }{ 100 } ) ^{11}\]

Answer 5

would i multiplu 100 to both sides ?

Answer 6

divide by 100000 on both sides

Answer 7

1.96

Answer 8

you probably should be using the continuous interest formula
http://cs.selu.edu/~rbyrd/math/continuous/

\[ P= P_0 e^{r \ t} \]

Answer 9

\[1.96 = (1+\frac{ r }{ 100 })^{11}\]

Answer 10

now can i divide by 100

Answer 11

no you need to take log

Answer 12

if you assume compounding at a yearly rate, your answer will be off by a few tenths of a percent

Answer 13

yea it asked for yearly rate of appreciation

Answer 14

log1.96=log (1+r/100)^2

Answer 15

^11****

Answer 16

no

Answer 17

how can there be power after you log

Answer 18

im confused

Answer 19

\[\log_{11} (1 + \frac{ r }{ 100 }) = 1.96 \]

Answer 20

\[\log_{11} (1.96) = 0.28064\]

Answer 21

***Assuming a steady rate of growth*** means continuous

they want the annual interest rate assuming continuous compounding.

Answer 22

yea thats what we've been using

Answer 23

okay i about to solve it tell me if im correct

Answer 24

2.8%?

Answer 25

I believe you are to assume annual compounding. By "steady rate of growth" I believe the author simply means that the annual rate of appreciation is constant over the duration of the problem.

Answer 26

no not 2.8

Answer 27

dayum if thats the case its simple interest formula

Answer 28

you move the decimal two places forward or back

Answer 29

And base 11 logs are not required. You have\[1.96=\left( 1+\frac{ r }{ 100 } \right)^{11}\]Using base 10 logs\[\log1.96 = 11 \log \left( 1+\frac{ r }{ 100 } \right)\]\[\frac{ \log 1.96 }{ 11 } = \log \left( 1+\frac{ r }{ 100 } \right)\]\[0.02657 = \log \left( 1+\frac{ r }{ 100 } \right)\]\[1+\frac{ r }{ 100 } = 10^{0.02657} = 1.0631\]Can you take it from here and solve for \(r\)?

Answer 30

no i dont understand

Answer 31

would it be .0631/100 ?

Answer 32

yea

Answer 33

0.000631 = r

Answer 34

So close. \[1+\frac{ r }{ 100 } = 0.0631\]Subtract 1 from both sides\[\frac{ r }{ 100 } = 0.0631\]How do you solve for r?

Answer 35

6.31

Answer 36

Sorry. First line should read\[1+\frac{ r }{ 100 } = 1.0631\]

Answer 37

6.31%

Answer 38

You are correct. Now round to the number of decimal places required. Well done?

Answer 39

Well done!

Answer 40

now you need to round off somemore

Answer 41

can yall help me with another ?

Answer 42

Delete this question and post the other.

Answer 43

it said that was wrong . it says to round to the nearest tenth of percent

Answer 44

6.3% correct ??

Answer 45

yea

Answer 46

i told you to round off somemore lol

Answer 47

it still says its wrong

Answer 48

We both indicated that you were to round to the correct number of decimal places :)

Answer 49

its still wrong . i put 6.3%

Answer 50

Well. I don't know what the issue is. \[196000 = 100000\times 1.603^{11}\]so the math checks out.

Answer 51

Sorry. 1.063^11

Answer 52

wanna try simple interest method?

Answer 53

I can't imagine that it's continuous compounding but, if so, you'll have to use natural logs.\[F=Pe^{rt}\]\[196000 = 100000e^{11r}\]\[1.96 = e^{11r}\]\[\ln 1.96 = 11r\]\[r=\frac{ \ln 1.96 }{ 11 }\]

Answer 54

okay .

Answer 55

.06117

Answer 56

Convert to percent by multiplying by 100% and round to the correct number of decimal places.

Answer 57

hate that kinda english-testing maths questions

Answer 58

yes, you get 6.1%

Answer 59

Me too. Just finished a graduate course in financial mathematics and "steady rate of growth" was never used to indicate continuous compounding. @phi , if this is the correct answer, I owe you an apology. Please forgive my doubt.

Answer 60

it was correct

Answer 61

Congratualtions

Answer 62

sorta can't understand how e comes into play in continuous compounding

Answer 63

gratz

Answer 64

THANK YALL !!!! .


can yall help me with the next one