An amount of $100 being increased every day by 1%, in how many days will it reach $1000 and how did you calculate this?

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  • Roger
    Lv 7
    4 weeks ago

    $1,000 = $100(1+ 0.01)^d where d is the number of days

    10 = (1.01)^d

    Log 10 = d(log 1.01)

    1 = d(0.00432)

    d = 231 days

  • 4 weeks ago

    An increase of 1% is the same as multiplying by 1.01, so the value after n days is:

    A_n = 100 * 1.01^n

    You want to find the value of n that makes A_n equal to 1000:

    100 * 1.01^n = 1000

    1.01 ^ n = 10                        Divide both sides by 100

    log (1.01^n) = log 10           Take logarithms

    n * (log 1.01) = log 10          Use log (a^b) = b * log a property

    n = (log 10) / (log 1.01)        ...and divide by (log 1.01)

    That works for any logarithm base, so long as you use the same one every time.  If "log" is a base ten ("common") logarithm then that simplifies to 1/(log 1.01).

    My calculator says that makes n about 231.41, so the first time the amount gets to $1000 or more is at the end of the 232nd day.

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