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Math Help - exponential

  1. #1
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    exponential

    I'm having a horrible problem with logarithms...rather natural logarithms. Okay so here's the problem.

    5(4^(2x-1))=21

    Now I have the soloution, but I need to understand how they got to the solution. There is one other problem, and I'm not sure about it....

    But that's a posting for after we work on this one.

    PLEASE HELP!!!!
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  2. #2
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    dont scream at me if i got it wrong or haven't answer ur question

    5(4^(2x-1))=21

    4^(2x-1)=21 / 5
    =4.2

    log[4^(2x-1)]=log4.2

    Rule: log (a^x)=x loga

    (2x-1) log4 = log4.2

    2x-1 = log4.2 / log 4

    x= [(log 4.2 / log 4) + 1] / 2

    that should be the answer!
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by calculus_jy View Post
    5(4^(2x-1))=21

    4^(2x-1)=21 / 5
    =4.2

    log[4^(2x-1)]=log4.2

    Rule: log (a^x)=x loga

    (2x-1) log4 = log4.2

    2x-1 = log4.2 / log 4

    x= [(log 4.2 / log 4) + 1] / 2

    that should be the answer!
    Looks good to me. I could scream that...

    -Dan
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  4. #4
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    Thank you guys but....

    See I knew I wasn't crazy I don't understand this at all....The answer is

    ln21-ln5+ln4
    2ln4


    I have no idea. I got something with Log also so I know I'm not crazy.

    Infact I got the exact thing you did.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DEvans View Post
    See I knew I wasn't crazy I don't understand this at all....The answer is

    ln21-ln5+ln4
    2ln4


    I have no idea. I got something with Log also so I know I'm not crazy.

    Infact I got the exact thing you did.
    First, you can use the log to any base you like, so you both have it right.

    Also note that
    log(4.2) = log(21/5) = log(21) - log(5)
    (again, the log to any base) so this also works.

    -Dan
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  6. #6
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    should be rite



    ln21-ln5+ln4
    2ln4


    here is how to get that

    5[4^(2x+1)] = 21

    4^(2x+1) = 21/5

    ln[4^(2x+1)] = ln (21/5)

    Rule:
    log(a/b)= loga - logb

    (2x+1)ln4 = (ln21 - ln 5)

    (2x+1) = (ln21 - ln5) / ln4

    2x = (ln21 - ln 5) / ln4 +1= (ln21 - ln5 + ln4) / ln4

    x= ln21-ln5+ln4
    2ln4
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  7. #7
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    I've got it

    I got confused thinking log was what I needed but if I start with a ln after dividing 21 by 5 then this is the result

    ln(4^(2x-1)) = ln(21/5)

    pull down (2x-1)

    (2x-1)ln4 = ln(21/5)

    divide by ln4

    2x-1 = ln(21/5)/ln4

    separate ln(21/5) into (ln21 - ln5)

    2x-1 = (ln21 - ln5)/ln4

    add 1

    2x = [(ln21-ln5)+ln4]/ln4

    divide by 2

    x = (ln21-ln5+ln4)/2ln4
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  8. #8
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    The other problem..

    And it should be simple enough. I'm just not sure if I understand what's being asked.

    Find the rule for the inverse of f if f(x) =

    What do you guys think?
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  9. #9
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    think this is rite

    f(x) =

    inverse function of f(x) is donated by f^-1 (x)

    this operation is to isolate f^-1 (x) when x is replaced by f^-1 (x) and f(x) by x in the original function

    thus

    x = (2 . f^-1 (x) - 3) / ( f^-1 (x) - 5 )
    x . f^-1 (x) - 5x = 2 . f^-1 (x) - 3
    (x - 2) . f^-1 (x) = -3 + 5x
    f^-1 (x) = (-3 + 5x) / (x - 2)

    This should be rite!
    Last edited by calculus_jy; April 30th 2007 at 08:55 PM. Reason: typed wrong
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  10. #10
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    Thanks I ended up getting it, what I needed to do was to subtract xy after I got the bottom cleared. Then after that factor it out. I was missing that vital step.

    Thanks alot guys you've been a big help.
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  11. #11
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DEvans View Post
    And it should be simple enough. I'm just not sure if I understand what's being asked.

    Find the rule for the inverse of f if f(x) =

    What do you guys think?
    If you have a new question, please place it in a new thread.

    -Dan
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