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Math Help - Log question

  1. #1
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    Log question

    Hi,
    This is a log problem that I can't solve..... I am kinda new to logs so any details (name of rule used....etc) would be very much appreciated. :-)

    2(5^x+1) = 1 + 3/5^x

    Thanks in advance
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  2. #2
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    Re: Log question

    Replace 5^x with y and solve the equation for y. Then solve 5^x = y for the found value of y.
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  3. #3
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    Re: Log question

    Thanks... but can you please show me how to get a value for x, I constructed the equation you told me to.... but what do I do now???
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  4. #4
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    Re: Log question

    Quote Originally Posted by IBstudent View Post
    Thanks... but can you please show me how to get a value for x, I constructed the equation you told me to.... but what do I do now???
    Post some working if you wish to be assisted further.
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  5. #5
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    Re: Log question

    Quote Originally Posted by Quacky View Post
    Post some working if you wish to be assisted further.
    fine....
    I am sure it is 100% wrong, but since you insist.....
    2(5^x+1) = 1 + 3/y
    2(5^x+1) = y + 3/y
    2y (5^x+1) = y+3
    y( 5^x+1) = 3
    5^2x+1 = 3 (I have a feeling I messed up bigtime, but I'll continue for your sake...)
    (2x+1) log5 = log3
    2x(log5) + log5 = log3
    (stuck)

    can you please help me now?


    Thanks
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  6. #6
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    Re: Log question

    I recommend replacing both occurrences of 5^x with y in the original equation and solving the result for y.
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  7. #7
    Super Member Quacky's Avatar
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    Re: Log question

    Quote Originally Posted by IBstudent View Post
    fine....
    I am sure it is 100% wrong, but since you insist.....
    2(5^x+1) = 1 + 3/y
    2(5^x+1) = y + 3/y
    2y (5^x+1) = y+3
    y( 5^x+1) = 3
    5^2x+1 = 3 (I have a feeling I messed up bigtime, but I'll continue for your sake...)
    (2x+1) log5 = log3
    2x(log5) + log5 = log3
    (stuck)

    can you please help me now?


    Thanks
    As emakarov said, remember that you let 5^x=y? You can do that here too:
    2(5^x+1) = 1 + \frac{3}{y}
    2(y+1)=1+\frac{3}{y}
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  8. #8
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    Re: Log question

    Quote Originally Posted by Quacky View Post
    As emakarov said, remember that you let 5^x=y? You can do that here too:
    2(5^x+1) = 1 + \frac{3}{y}
    2(y+1)=1+\frac{3}{y}
    ok so:
    2(y+1) = 1 + 3/y
    2y+2 = 1+ 3/y
    2y+1 = 3/y
    cross multiplying
    2y^2+y=3
    2y^2+y-3=0
    aha! a quadratic equation!
    solving it using -b+-sqrt(b^2-4ac)/2a gives either 1 or -1.5. keeping in mind that this is 5^x. x could be 0 or.... undefined...
    unfortunately, this is not the answer (not even close for that matter) :-(

    guys, please, would it kill you to answer the question and show your steps...?

    Thanks
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  9. #9
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    Re: Log question

    Your solution is correct: x = 0 is the only solution of this equation. You can check that it is a solution by substituting 0 for x in the original equation.
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  10. #10
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    Re: Log question

    sustituting gives 10=4
    the mark scheme says that x is equivalent to -1 + log5(3) (by 5 I mean the base)
    how is this possible?
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  11. #11
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    Re: Log question

    2(5^x+1) = 1 + 3/5^x
    Left-hand side: 2(5^0 + 1) = 2(1 + 1) = 2 * 2 = 4.
    Right-hand side: 1 + 3/5^0 = 1 + 3/1 = 1 + 3 = 4.
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  12. #12
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    Re: Log question

    oh I see thanks......
    I guess the mark scheme is wrong and we are right!
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  13. #13
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    Re: Log question

    On the other hand, the mark scheme is right if the equation is

    \displaystyle2\cdot5^{x+1} = 1 + \frac{3}{5^x},

    which in ASCII can be represented as

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

    Then

    2\cdot 5^{x+1}=2\cdot5\cdot 5^x=10\cdot5^x.

    Replacing 5^x by y again and solving for y gives y = 3/5. So, 5^x=3/5, or 5^{x+1}=3, from where x+1=\log_53. Alternatively, 5^x=3/5 gives

    x = \log_5(3/5) = \log_53-\log_55 = \log_53-1.
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  14. #14
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    Re: Log question

    Last question....
    so I know that x is -0.317394....... but, how did you know that this value+1 is equal to log5(3). It that a log rule? please tell me your method.
    Thanks for your help :-)
    Last edited by IBstudent; June 22nd 2011 at 07:49 AM.
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  15. #15
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    Re: Log question

    I did not claim above that \log_53\approx −0.317394, though this is true. If you don't have a calculator that computes logarithms to an arbitrary base, you can use the following equality.

    \log_ab\log_bc=\log_ac (1)

    In particular,

    \log_53=\frac{\log_e3}{\log_e5}=\frac{\ln3}{\ln5}. (2)

    The mnemonic for (1) is transitivity: (a,b) and (b,c) gives you (a,c). For (2), the argument of ln() in the nominator becomes the argument of log() and the argument from the denominator becomes the base, which is located in a subscript below the log argument, just like the denominator is below the nominator.

    Also, see Wikipedia.

    Starting from 5^x=3/5, you could also take ln() of both sides to get \ln(5^x)=x\ln 5=\ln(3/5)=\ln3-\ln5, from where (x+1)\ln5=\ln3 and x=-1+\frac{\ln3}{\ln5}.

    Edit: I am not sure I addressed the question you asked. If not, feel free to rephrase it.
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