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A couple more

  1. Sep 13, 2006 #1
    This is not my day.

    Solve for x.

    1.) 9^2x+1 = 27

    2.) 10^5x = 1000


    Thanks again!
     
  2. jcsd
  3. Sep 14, 2006 #2
    assuming base 2:
    log30=4x is equivalent to writing 2^4x=30
    log30 is equivalent to writing (log30)/(log2)
    This means (log30)/(log2)=4x

    assuming base 8:
    log12=5x+4 is equivalent to writing 8^(5x+4)=12
    log12 is equivalent to writing (log12)/(log8)
    This means (log12)/(log8)=5x+4


    I hope those hints are enough. If you still can't figure it out just ask
     
  4. Sep 14, 2006 #3

    J77

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    You can also think about equating exponents:

    Just to clarify, but in your first question, you must mean 9^(2x+1)=27

    As an example,

    4^(3x-1)=8

    is equiv. to

    2^(2(3x-1))=2^3

    equating exponents:

    6x-2=3

    which implies:

    x=5/4

    Now do the same for your problems...
     
  5. Sep 14, 2006 #4

    VietDao29

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    Homework Helper

    You should use logarithm to solve the problems above.
    Say if you need to solve ax = b for x, where a and b are 2 constants. Then you can take loga both sides, like this:
    ax = b
    <=> loga(ax) = logab
    <=> x = logab
    I'll give you an example:
    52x = 25
    You can do this in 2 ways:
    First way:
    52x = 25 = 52
    <=> 2x = 2
    <=> x = 1
    Second way:
    52x = 25
    <=> log5(52x) = log5(25)
    <=> 2x = 2
    <=> x = 1
    Ok, can you do the problems? Just try it again and see. :)
     
    Last edited: Sep 14, 2006
  6. Sep 14, 2006 #5

    J77

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    Did you not just repeat the two methods which me and steve gave? :confused:
     
  7. Sep 14, 2006 #6

    VietDao29

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    Homework Helper

    Whoops, I am sorry, didn't pay attention when posting it. :frown:
    Please don't take it personally. :smile: o:)
     
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