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)

- anonymous

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- anonymous

@saseal

- anonymous

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

- anonymous

yea

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## More answers

- anonymous

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

- anonymous

would i multiplu 100 to both sides ?

- anonymous

divide by 100000 on both sides

- anonymous

1.96

- phi

you probably should be using the continuous interest formula
http://cs.selu.edu/~rbyrd/math/continuous/
\[ P= P_0 e^{r \ t} \]

- anonymous

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

- anonymous

now can i divide by 100

- anonymous

no you need to take log

- phi

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

- anonymous

yea it asked for yearly rate of appreciation

- anonymous

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

- anonymous

^11****

- anonymous

no

- anonymous

how can there be power after you log

- anonymous

im confused

- anonymous

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

- anonymous

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

- phi

***Assuming a steady rate of growth*** means continuous
they want the annual interest rate assuming continuous compounding.

- anonymous

yea thats what we've been using

- anonymous

okay i about to solve it tell me if im correct

- anonymous

2.8%?

- anonymous

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.

- anonymous

no not 2.8

- anonymous

dayum if thats the case its simple interest formula

- anonymous

you move the decimal two places forward or back

- anonymous

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 \(r\)?

- anonymous

no i dont understand

- anonymous

would it be .0631/100 ?

- anonymous

yea

- anonymous

0.000631 = r

- anonymous

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

- anonymous

6.31

- anonymous

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

- anonymous

6.31%

- anonymous

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

- anonymous

Well done!

- anonymous

now you need to round off somemore

- anonymous

can yall help me with another ?

- anonymous

Delete this question and post the other.

- anonymous

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

- anonymous

6.3% correct ??

- anonymous

yea

- anonymous

i told you to round off somemore lol

- anonymous

it still says its wrong

- anonymous

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

- anonymous

its still wrong . i put 6.3%

- anonymous

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

- anonymous

Sorry. 1.063^11

- anonymous

wanna try simple interest method?

- anonymous

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

- anonymous

okay .

- anonymous

.06117

- anonymous

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

- anonymous

hate that kinda english-testing maths questions

- phi

yes, you get 6.1%

- anonymous

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.

- anonymous

it was correct

- anonymous

Congratualtions

- anonymous

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

- anonymous

gratz

- anonymous

THANK YALL !!!! .
can yall help me with the next one

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