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Math Help - Exponents and logarithms

  1. #1
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    Exponents and logarithms

    I just joined this website because I was looking for a little advice/help with math work. There are a few questions I had for homework that I haven't been able to solve.

    I put the questions in an image to keep the formatting.
    I learn better by having the work done and shown step by step. I'd prefer it to not have the actual equations solved, but ones similar to them, because I wish to solve them myself.
    Thank you.

    Equations:
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by bnosam View Post
    I just joined this website because I was looking for a little advice/help with math work. There are a few questions I had for homework that I haven't been able to solve.

    I put the questions in an image to keep the formatting.
    I learn better by having the work done and shown step by step. I'd prefer it to not have the actual equations solved, but ones similar to them, because I wish to solve them myself.
    Thank you.

    Equations:
    a) Example: 2^{3x-1} = 11^{x+2}

    Use the log power law: (3x-1)\ln(2) = (x+2)\ln(11)

    Distribute 3\ln(2)x - \ln(2) = x\ln(11) + 2\ln(11)

    Simplify 3\ln(2)x - x\ln(11) = 2\ln(11) + \ln(2)

    Factorise x(3\ln(2)-\ln(11)) = 2\ln(11) + \ln(2)

    Solve x = \frac{2\ln(11) + \ln(2)}{3\ln(2)-\ln(11)}


    -----------------------------------

    b) 4 \cdot 4^{x} = 5

    Since the base is the same there are two ways to do this:

    1)Laws of exponents say that 4 \cdot 4^x = 4^{x+1}

    Then the same as question a)

    2) 4^x = \frac{5}{4}

    take logs: x\ln(4) = \ln(5)-\ln(4)

    x = \frac{\ln(5)-\ln(4)}{\ln(4)}


    ----------------------------

    c) Treat as a quadratic equation: u = \log_3(x)

    3u^2 -28u+9 =0

    Solve using the quadratic formula but remember that \log_3(x) > 0
    Last edited by e^(i*pi); May 8th 2010 at 01:26 PM. Reason: latex
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  3. #3
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    Quote Originally Posted by e^(i*pi) View Post
    a) Example: 2^{3x-1} = 11^{x+2}

    Use the log power law: (3x-1)\ln(2) = (x+2)\ln(11)

    Distribute 3\ln(2)x - \ln(2) = x\ln(11) + 2\ln(11)

    Simplify 3\ln(2)x - x\ln(11) = 2\ln(11) + \ln(2)

    Factorise x(3\ln(2)-\ln(11)) = 2\ln(11) + \ln(2)

    Solve x = \frac{2\ln(11) + \ln(2)}{3\ln(2)-\ln(11)}


    -----------------------------------

    b) 4 \cdot 4^{x} = 5

    Since the base is the same there are two ways to do this:

    1)Laws of exponents say that 4 \cdot 4^x = 4^{x+1}

    Then the same as question a)

    2) 4^x = \frac{5}{4}

    take logs: x\ln(4) = \ln(5)-\ln(4)

    x = \frac{\ln(5)-\ln(4)}{\ln(4)}


    ----------------------------

    c) Treat as a quadratic equation: u = \log_3(x)

    3u^2 -28u+9 =0

    Solve using the quadratic formula but remember that \log_3(x) > 0
    I don't mean to take away from anything that you've said, but in my math class, we haven't discussed what "ln" means or what it does. So, using it would probably just confuse me more, I think.
    Thank you though.

    Edit: Just saw that you were an Iron Maiden fan in your profile. Good call on that!
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by bnosam View Post
    I don't mean to take away from anything that you've said, but in my math class, we haven't discussed what "ln" means or what it does. So, using it would probably just confuse me more, I think.
    Thank you though.

    Edit: Just saw that you were an Iron Maiden fan in your profile. Good call on that!
    ln is a special name for the logarithm with base e.

    \ln(x) = \log_e(x)

    There is no reason why you can't use any other base such as base 10


    ============================


    Edit: Same working as above but using base 10 (find and replace <3):


    Example: 2^{3x-1} = 11^{x+2}

    Use the log power law: (3x-1)\log_{10}(2) = (x+2)\log_{10}(11)

    Distribute 3\log_{10}(2)x - \log_{10}(2) = x\log_{10}(11) + 2\log_{10}(11)

    Simplify 3\log_{10}(2)x - x\log_{10}(11) = 2\log_{10}(11) + \log_{10}(2)

    Factorise x(3\log_{10}(2)-\log_{10}(11)) = 2\log_{10}(11) + \log_{10}(2)

    Solve x = \frac{2\log_{10}(11) + \log_{10}(2)}{3\log_{10}(2)-\log_{10}(11)}


    -----------------------------------

    b) 4 \cdot 4^{x} = 5

    Since the base is the same there are two ways to do this:

    1)Laws of exponents say that 4 \cdot 4^x = 4^{x+1}

    Then the same as question a)

    2) 4^x = \frac{5}{4}

    take logs: x\log_{10}(4) = \log_{10}(5)-\log_{10}(4)

    x = \frac{\log_{10}(5)-\log_{10}(4)}{\log_{10}(4)}
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