from time t=0, with t in years, a $1200 deposit in a bandkaccount grows accorkding to the formulaB=1200(exponential)^(0.03t) a)what is the balance in the account at the of 100years? b)when does the balance first go over $50,000?

Matonya

Matonya

Answered question

2022-08-04

From time t=0, with t in years, a $1200 deposit in a bandkaccount grows accorkding to the formula B=1200(exponential)0.03t
a)what is the balance in the account at the of 100years?
b)when does the balance first go over $50,000?

Answer & Explanation

Kasen Schroeder

Kasen Schroeder

Beginner2022-08-05Added 21 answers

B = 1200 e 0.03 t
a) for this part, t = 100, so we substitute this into the equation andsolve:
B = 1200 e 0.03 100 = 1200 e 3 24 , 102.6 solved this using a calculator, sothe balance at 100 years is approximately $24,102.6
b) for this part, the balance or B is greater than 50,000, so settingup the equation
50 , 000 < 1200 e 0.03 t divide both sides by 1200
50 , 000 1200 < 0.03 t take the ln of both sides
ln ( 50 , 000 1200 ) < 0.03 t divide everything by 0.03
ln ( 50 , 000 1200 ) 0.03 < t
t> 124.323
So the balance will go over $50,000 after about 124.323 years
musicintimeln

musicintimeln

Beginner2022-08-06Added 3 answers

a)what is the balance in the account at the of 100years?
When t = 100
B = 1200 e 0.03 t = 1200 e 0.03 ( 100 ) = 1200 e 3 = $ 24102.6
b)when does the balance first go over $50,000?
B = 50,000 and solve for t
50000 = 1200 e 0.03 t
50000 ÷ 1200 = e 0.03 t
125 / 3 = e 0.03 t
Take natural log on both sides to elimitate e
ln ( 125 / 3 ) = 0.03 t
t = ln ( 125 / 3 ) / 0.03
t=124.32
Therefore, the balance goes over $50 000 after about 124years and 4 months

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