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Thread: Please Help

  1. #1
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    Please Help

    Is A*B^(x+1) = A*(x+1)*lnB ?

    if not how could I find x?
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  2. #2
    Super Member bigwave's Avatar
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    $\displaystyle AB^{(x+1)} = A(x+1)lnB $

    divide both sides by A

    $\displaystyle B^{(x+1)} = (x+1)lnB \Rightarrow B^{(x+1)} = lnB^{(x+1)}$

    if (x+1) = 0 then x = -1

    so

    $\displaystyle B^{0} = lnB^{0} \Rightarrow 1 = 1$
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  3. #3
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    consider the case where A,B and x all equal 1.
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  4. #4
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    Quote Originally Posted by bigwave View Post
    $\displaystyle AB^{(x+1)} = A(x+1)lnB $

    divide both sides by A

    $\displaystyle B^{(x+1)} = (x+1)lnB \Rightarrow B^{(x+1)} = lnB^{(x+1)}$

    do you know what x has to be so these will be equal
    I'm sorry, I forgot: let's say it's =2
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  5. #5
    Super Member bigwave's Avatar
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    if the exponents are 0 the term goes to 1

    try

    if $\displaystyle (x+1) = 0$
    then $\displaystyle x = -1$

    if the exponents are 0 the terms goes to 1

    also check out pickslides question
    Last edited by bigwave; Jan 6th 2010 at 12:59 PM. Reason: clean up
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  6. #6
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    Quote Originally Posted by bigwave View Post
    try

    if $\displaystyle (x+1) = 0$
    then $\displaystyle x = -1$

    if the exponents are 0 the terms goes to 1

    also check out pickslides question
    I know X equals 75 (more or less)
    The equation I need to solve s this:
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