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Thread: Logarithm simplification (probably easy)

  1. #1
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    Logarithm simplification (probably easy)

    Ok so as the title says, it's probably really easy but I'm having trouble getting a value of x:

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

    I've got down to

    xlog(4) = log(2^(x+2)-3)

    And I have NO idea how to simplify to get rid of that -3.... Please help! I'm sure there's something REALLY obvious that I'm missing Thank you!
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    Quote Originally Posted by Natalie11391 View Post
    Ok so as the title says, it's probably really easy but I'm having trouble getting a value of x:

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

    I've got down to

    xlog(4) = log(2^(x+2)-3)

    And I have NO idea how to simplify to get rid of that -3.... Please help! I'm sure there's something REALLY obvious that I'm missing Thank you!
    Already answered below
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  3. #3
    MHF Contributor Unknown008's Avatar
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    You don't need to use logs... well not now.

    Make a substitution:

    y = 2^x

    How about it?
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    Quote Originally Posted by Unknown008 View Post
    You don't need to use logs... well not now.

    Make a substitution:

    y = 2^x

    How about it?
    You wouldn't believe how hard I'm kicking myself now.

    To the OP, when working out your solutions remember that for all real x: 2^x > 0
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  5. #5
    MHF Contributor Unknown008's Avatar
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    Oh, I think I know how you feel, I felt the same thing before getting the idea for the substitution.

    Note to OP: Sometimes, taking logs immediately only make things harder. Look at the bases first.
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    Thanks for your help, but I still can't do it!! It must be lack of sleep but I can't substitute 2^x as y for 4^x

    I know that 2^2 = 4 so 2^2x = 4^x but then I can't seem to get my head around what to do next!
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  7. #7
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    Quote Originally Posted by Natalie11391 View Post
    Thanks for your help, but I still can't do it!! It must be lack of sleep but I can't substitute 2^x as y for 4^x

    I know that 2^2 = 4 so 2^2x = 4^x but then I can't seem to get my head around what to do next!
    That's right so sub in 2^{2x} for 4^x

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

    2^{2x} +3 = 2^{x+2}

    You also know that 2^{x+2} = 4 \cdot 2^x (which is the reverse of what you did in the OP)

    2^{2x} + 3 = 4 \cdot 2^x and if you rearrange it you get (2^x)^2 - 4 \cdot 2^x + 3 = 0

    That is a quadratic equation in 2^x so solve using your favourite method
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