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Math Help - Logarithms help

  1. #1
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    Logarithms help

    Well I have 2 questions but they are labeled (i) and (ii) from the same question so I'll post them in a single thread. Please forgive me if I'm not supposed to that because I just joined today and I'm not too familar with this forum.
    I don't understand how to type in latex so I will attach pictures.



    These are the steps I used for part (i):

    And for part (ii):


    The answer key says x = 2 which I do not quite understand because there are two part questions, so I assume it is wrong. My exams are coming soon and I find that I often come across this kind of problem so any help would be much appreciated, thanks!
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  2. #2
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    Quote Originally Posted by caramelcake View Post
    Well I have 2 questions but they are labeled (i) and (ii) from the same question so I'll post them in a single thread. Please forgive me if I'm not supposed to that because I just joined today and I'm not too familar with this forum.
    I don't understand how to type in latex so I will attach pictures.



    These are the steps I used for part (i):

    And for part (ii):


    The answer key says x = 2 which I do not quite understand because there are two part questions, so I assume it is wrong. My exams are coming soon and I find that I often come across this kind of problem so any help would be much appreciated, thanks!
    \log_2(x) + \log_8(2x) = \frac{7}{3}

    \log_8(2x) = \frac{\log_2(2x)}{\log_2(8)} = \frac{\log_2(2)+\log_2(x)}{3} = \frac{1+\log_2(x)}{3}


    \log_2(x) + \frac{1+\log_2(x)}{3} = \frac{7}{3}

    Multiply by 3

    3\log_2(x)+1+\log_2(x) = 7

    \log_2(x^3) + \log_2(x) = 6

    \log_2(x^4) = 6
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  3. #3
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    Thank you e^(i*pi) for your reply, but I do believe that your reply was exactly what I had already come to in my first post. Or is it that I can simply put \log_2(x^4) = 6 as my answer?
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  4. #4
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    \log_2{(x^4)} = 6

    x^4 = 2^6

    x = (2^6)^{\frac{1}{4}}

    x = 2^{\frac{3}{2}}

    x = 2\sqrt{2}.
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  5. #5
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    Thank you for your kind help! I didn't know that you could 'transfer' exponents like that, maybe that's why I'm always stuck when solving logarithm problems. Thanks again!
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  6. #6
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    \log{(4^x - 4)} - x\log{2} = \log{3}

    \log{(4^x - 4)} - \log{(2^x)} = \log{3}

    \log{\left(\frac{4^x - 4}{2^x}\right)} = \log{3}

    \frac{4^x - 4}{2^x} = 3

    \frac{(2^2)^{x} - 2^2}{2^x} = 3

    2^{2x} - 2^2 = 3(2^x)

    (2^x)^2 - 3(2^x) - 4 = 0

    (2^x - 4)(2^x + 1) = 0


    2^x - 4 = 0 or 2^x + 1 = 0.

    But since 2^x > 0 for all x, that means the only case that works is

    2^x - 4 = 0

    2^x - 2^2 = 0

    2^x = 2^2

    x = 2.
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  7. #7
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    Cased closed!

    Thank you e^(i*pi) and Prove It for your quick help, I really appreciate it!
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