A student borrows $10 000 at an interest rate of 4% per annum, compounded monthly. a) If the student pays the loan off in 10 years, how much does the student pay in interest? # of compounding periods/year=12 total # of compounding periods = 12 x 10 = 120 Interest rate/compounding period = 4/12 = .33 Formula = 10,000 (1.0033)^120 =14849 = $4,849 in interest was paid in interest b) If the entire loan costs the student $20 000 to pay back, how many years was the loan for?

Intermediate Algebra
10th Edition
ISBN:9781285195728
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Jerome E. Kaufmann, Karen L. Schwitters
Chapter11: Exponential And Logarithmic Functions
Section11.2: Applications Of Exponential Functions
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Studying for my final and not sure if I am understanding this right

A student borrows $10 000 at an interest rate of 4% per annum, compounded monthly.
a) If the student pays the loan off in 10 years, how much does the student pay in interest?
# of compounding periods/year=12
total # of compounding periods = 12 x 10 = 120
Interest rate/compounding period = 4/12 = 33
Formula = 10,000 (1.0033) ^ 120
= 14849 = $4,849 in interest was paid in interest
b) If the entire loan costs the student $20 000 to pay back, how many years was the loan for?
ANON
20 000 = 10,000 (1.0033)
10 000 10,000
N=210.4
= 17.54 years.
2 = (1.0033) ^ Non
log 2 = Log|1.0033N
10g 220110g (1.303
21
log
12 Log 1.0033
9
Transcribed Image Text:A student borrows $10 000 at an interest rate of 4% per annum, compounded monthly. a) If the student pays the loan off in 10 years, how much does the student pay in interest? # of compounding periods/year=12 total # of compounding periods = 12 x 10 = 120 Interest rate/compounding period = 4/12 = 33 Formula = 10,000 (1.0033) ^ 120 = 14849 = $4,849 in interest was paid in interest b) If the entire loan costs the student $20 000 to pay back, how many years was the loan for? ANON 20 000 = 10,000 (1.0033) 10 000 10,000 N=210.4 = 17.54 years. 2 = (1.0033) ^ Non log 2 = Log|1.0033N 10g 220110g (1.303 21 log 12 Log 1.0033 9
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