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Never done logarithms

  1. Jan 13, 2004 #1
    IS there a way to find x if given:

    3^x - 3^(x-1) = 1000


    I tried to take log [3^x - 3^(x-1)]/log[3] = log1000/log3

    but then, x-x-1 = ...
     
  2. jcsd
  3. Jan 13, 2004 #2
    well, I've never done logarithms really, but I figured out how to do this


    to break it down it goes like this

    (3*3*3*3...*3*3) - (3*3*3*3..*3) = 1000

    so therefore 1 less 3 is multiplied in the second part...heres a simple example so you can see the pattern

    (5*5*5*5) - (5*5*5) = (5*5*5)*(5-1)

    therefore we can say in general that this means

    N^m - N^(m-1) = (N^(m-1))*(N-1)

    now instead of the left side of the equation above, we use the right side, and solve for x

    (3^(x-1))*(3-1) = 1000
    3^(x-1) = 1000/(3-1)
    3^(x-1) = 500
    log (3^(x-1)) = log (500)
    (x-1)(log 3) = log (500)
    x-1 = log (500)/log (3)
    x = [log (500)/log (3)] + 1

    and voila, you have x...

    another way of looking at this is that since you are subtracting something with one less *3, then 3^x : 3^(x-1) is 3:1. 3-1=2, so 2:1000. This means that 3:1500 and 1:500. Therefore 3^x = 1500 and x^(x-1) = 500

    I hope this helps
     
  4. Jan 14, 2004 #3

    Integral

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    You can use the properties of exponents to factor your inital expression.

    [tex] 3^x - 3^{(x-1)}=1000 [/tex]
    [tex]3^x -3^x 3^{-1}=1000 [/tex]
    [tex]3^x(1 - 3^{-1})=1000[/tex]
    [tex]3^x= 1500[/tex]
    [tex]ln(3^x)= ln 1500 [/tex]
    [tex]x ln 3= ln 1500 [/tex]
    [tex] x = \frac {ln 1500} {ln 3} [/tex]

    I'll leave it to the reader to show that the 2 results are the same.
     
  5. Jan 18, 2004 #4
    another solution:
    instead of factoring 3^x, factor out 3^x-1
    (3^x)-{3(x-1)}=1000
    {3^(x-1)}[3-1]=1000
    {3^(x-1)}2=1000
    {3^(x-1)}=500
    log{3^(x-1)}=log(500)
    (x-1)log3=log500
    x=[(log500)/(log3)]+1
     
  6. Nov 20, 2004 #5
    How can I understand the logarithms requirements in flowcharts? Easily please, I am mad about a situation with a proffesor.
     
  7. Nov 20, 2004 #6
    u mean something like

    log (a*b) = log a + log b?

    they are called Log Rules or Log Properties.
     
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