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Math Help - Solve Logarithm

  1. #1
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    Solve Logarithm

    I'm having the same kind of problems on all of the following problems: I cannot seem to isolate X. Help if you can!

    1. 5^(x+1)=e^x
    2. 3^(x+1)=(9^x)(27^(x+1))
    3. 10^(2x+1)=4^(x+1)
    4. 5^(3x-2)=3^(x+1)
    5. *Log Base is 2* Log(x+4) + Log(x+2)=Log3

    Much appreciated!
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  2. #2
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    Quote Originally Posted by ellimist121 View Post
    1. 5^(x+1)=e^x
    Take \ln of both sides to get, \ln 5^{x+1} = \ln e^x. Using identities we get (x+1) \ln 5 = x \ln e. But \ln e = 1. Thus, we are left with (x+1)\ln 5 = x \implies x \ln 5 + \ln 5 = x. Rewrite as x\ln 5 - x = - \ln 5. Factor, x(\ln 5 - 1) = - \ln 5. And this gives x = - \tfrac{\ln 5}{\ln 5 -1} = \tfrac{\ln 5}{1-\ln 5}

    2. 3^(x+1)=(9^x)(27^(x+1))
    This is,
    3^x = 3^{2x} \cdot 3^{3(x+1)} \implies 3^x = 3^{2x + 3(x+1)} \implies x = 2x + 3(x+1)
    Now solve for x

    Try doing the other problems.
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  3. #3
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    Thanks, but I'm still having trouble with the last one I have gotten to this:

    Log(((x+4)(x+2))/3)=0

    Also, with this one:

    (5^2x)-7(5^x)+12=0.
    Last edited by ellimist121; October 28th 2008 at 01:25 PM. Reason: advancement in problem
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  4. #4
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by ellimist121 View Post
    Thanks, but I'm still having trouble with the last one I have gotten to this:

    Log(((x+4)(x+2))/3)=0

    Also, with this one:

    (5^2x)-7(5^x)+12=0.
    2^0 = \frac{x^2+6x+8}{3}
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