Exponential Growth Algebra Question! HELP PLEASE!
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Exponential Growth Algebra Question! HELP PLEASE!

[From: ] [author: ] [Date: 12-05-27] [Hit: ]
031^(26)*(128) = $283.10 after 26 years)-280 = 128(1.log(280) = log(128) + y(log1.Then solve for y and put numbers in from the calculator for the logs and calculate the answer.......
I have $128 in my savings account. The bank pays me 3.1% annual interest. If I do not deposit or withdraw any money from the account, after how many years will my account balance first exceed $280? I would love some help on this! THANKS! =]

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The amount in the account is, after...

1 year: A_1 = 128 + 0.031(128) = 1.031(128)

2 years: A_2 = 1. 031^2(128)
...
n years: A_n = 1.031^n(128)

Then for A_n = 280 we have:

1.031^n(128) = 280

=> n*ln(1.031) = ln(280/128)

=> n = ln(280/128)/ln(1.031) = 25.6

So, since n has to be a whole number when n=26 years later you have at least $280 in the account. (or more precisely 1.031^(26)*(128) = $283.10 after 26 years)

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280 = 128(1.031)^y

Take logs

log(280) = log(128) + y(log1.031)

Then solve for y and put numbers in from the calculator for the logs and calculate the answer.
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