anonymous
  • anonymous
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)
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
anonymous
  • anonymous
growth : y = a(1+r)^2
anonymous
  • anonymous
yea

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anonymous
  • anonymous
\[196000 = 100000 (1 + \frac{ r }{ 100 } ) ^{11}\]
anonymous
  • anonymous
would i multiplu 100 to both sides ?
anonymous
  • anonymous
divide by 100000 on both sides
anonymous
  • anonymous
1.96
phi
  • phi
you probably should be using the continuous interest formula http://cs.selu.edu/~rbyrd/math/continuous/ \[ P= P_0 e^{r \ t} \]
anonymous
  • anonymous
\[1.96 = (1+\frac{ r }{ 100 })^{11}\]
anonymous
  • anonymous
now can i divide by 100
anonymous
  • anonymous
no you need to take log
phi
  • phi
if you assume compounding at a yearly rate, your answer will be off by a few tenths of a percent
anonymous
  • anonymous
yea it asked for yearly rate of appreciation
anonymous
  • anonymous
log1.96=log (1+r/100)^2
anonymous
  • anonymous
^11****
anonymous
  • anonymous
no
anonymous
  • anonymous
how can there be power after you log
anonymous
  • anonymous
im confused
anonymous
  • anonymous
\[\log_{11} (1 + \frac{ r }{ 100 }) = 1.96 \]
anonymous
  • anonymous
\[\log_{11} (1.96) = 0.28064\]
phi
  • phi
***Assuming a steady rate of growth*** means continuous they want the annual interest rate assuming continuous compounding.
anonymous
  • anonymous
yea thats what we've been using
anonymous
  • anonymous
okay i about to solve it tell me if im correct
anonymous
  • anonymous
2.8%?
anonymous
  • 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
  • anonymous
no not 2.8
anonymous
  • anonymous
dayum if thats the case its simple interest formula
anonymous
  • anonymous
you move the decimal two places forward or back
anonymous
  • 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
  • anonymous
no i dont understand
anonymous
  • anonymous
would it be .0631/100 ?
anonymous
  • anonymous
yea
anonymous
  • anonymous
0.000631 = r
anonymous
  • 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
  • anonymous
6.31
anonymous
  • anonymous
Sorry. First line should read\[1+\frac{ r }{ 100 } = 1.0631\]
anonymous
  • anonymous
6.31%
anonymous
  • anonymous
You are correct. Now round to the number of decimal places required. Well done?
anonymous
  • anonymous
Well done!
anonymous
  • anonymous
now you need to round off somemore
anonymous
  • anonymous
can yall help me with another ?
anonymous
  • anonymous
Delete this question and post the other.
anonymous
  • anonymous
it said that was wrong . it says to round to the nearest tenth of percent
anonymous
  • anonymous
6.3% correct ??
anonymous
  • anonymous
yea
anonymous
  • anonymous
i told you to round off somemore lol
anonymous
  • anonymous
it still says its wrong
anonymous
  • anonymous
We both indicated that you were to round to the correct number of decimal places :)
anonymous
  • anonymous
its still wrong . i put 6.3%
anonymous
  • anonymous
Well. I don't know what the issue is. \[196000 = 100000\times 1.603^{11}\]so the math checks out.
anonymous
  • anonymous
Sorry. 1.063^11
anonymous
  • anonymous
wanna try simple interest method?
anonymous
  • 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
  • anonymous
okay .
anonymous
  • anonymous
.06117
anonymous
  • anonymous
Convert to percent by multiplying by 100% and round to the correct number of decimal places.
anonymous
  • anonymous
hate that kinda english-testing maths questions
phi
  • phi
yes, you get 6.1%
anonymous
  • 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
  • anonymous
it was correct
anonymous
  • anonymous
Congratualtions
anonymous
  • anonymous
sorta can't understand how e comes into play in continuous compounding
anonymous
  • anonymous
gratz
anonymous
  • anonymous
THANK YALL !!!! . can yall help me with the next one

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