anonymous one year ago 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)

1. anonymous

@saseal

2. anonymous

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

3. anonymous

yea

4. anonymous

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

5. anonymous

would i multiplu 100 to both sides ?

6. anonymous

divide by 100000 on both sides

7. anonymous

1.96

8. phi

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

9. anonymous

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

10. anonymous

now can i divide by 100

11. anonymous

no you need to take log

12. phi

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

13. anonymous

yea it asked for yearly rate of appreciation

14. anonymous

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

15. anonymous

^11****

16. anonymous

no

17. anonymous

how can there be power after you log

18. anonymous

im confused

19. anonymous

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

20. anonymous

$\log_{11} (1.96) = 0.28064$

21. phi

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

22. anonymous

yea thats what we've been using

23. anonymous

okay i about to solve it tell me if im correct

24. anonymous

2.8%?

25. 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.

26. anonymous

no not 2.8

27. anonymous

dayum if thats the case its simple interest formula

28. anonymous

you move the decimal two places forward or back

29. 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$$?

30. anonymous

no i dont understand

31. anonymous

would it be .0631/100 ?

32. anonymous

yea

33. anonymous

0.000631 = r

34. 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?

35. anonymous

6.31

36. anonymous

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

37. anonymous

6.31%

38. anonymous

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

39. anonymous

Well done!

40. anonymous

now you need to round off somemore

41. anonymous

can yall help me with another ?

42. anonymous

Delete this question and post the other.

43. anonymous

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

44. anonymous

6.3% correct ??

45. anonymous

yea

46. anonymous

i told you to round off somemore lol

47. anonymous

it still says its wrong

48. anonymous

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

49. anonymous

its still wrong . i put 6.3%

50. anonymous

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

51. anonymous

Sorry. 1.063^11

52. anonymous

wanna try simple interest method?

53. 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 }$

54. anonymous

okay .

55. anonymous

.06117

56. anonymous

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

57. anonymous

hate that kinda english-testing maths questions

58. phi

yes, you get 6.1%

59. 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.

60. anonymous

it was correct

61. anonymous

Congratualtions

62. anonymous

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

63. anonymous

gratz

64. anonymous

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