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Math Help - Please clarify my workings for a somewhat complex (for me) logarithm question

  1. #1
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    Please clarify my workings for a somewhat complex (for me) logarithm question

    Hi all,

    I've been confronted with a somewhat complex logarithm question in the form of solving for x. After a long time of thinking and trying various methods, I finally came up with two possible methods for solving the problem. However, I am not sure which (or if either) is correct, or if I have made a mistake along the way. Please clarify the correct method.

    The question is:

    log2(X ^ log2(x)) = 4

    From there, I used the power rule and brought log2(x) down the front of the equation:

    log2(x) * log2(x) = 4

    Then I added the logs together:

    log2(x) + log2(x) = 4

    log2(2x) = 4

    Then I rearranged the equation to indical form:

    2 ^ 4 = 2X

    16 = 2X

    x = 8

    Thus, x = 8. However, I want to suggest an alternate method as well that returns to this part of the equation:

    log2(x) * log2(x) = 4

    Perhaps instead of adding them together like I did in the above working, I simply multiply the two logs together to achieve:

    log2(x ^ 2) = 4

    Rearrange and solve to equal:

    2 ^ 4 = x ^ 2

    16 = x ^ 2

    x = 4


    Any help will be much appreciated!

    Thanks,

    Nathaniel
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  2. #2
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    Quote Originally Posted by BinaryBoy View Post
    Hi all,

    I've been confronted with a somewhat complex logarithm question in the form of solving for x. After a long time of thinking and trying various methods, I finally came up with two possible methods for solving the problem. However, I am not sure which (or if either) is correct, or if I have made a mistake along the way. Please clarify the correct method.

    The question is:

    log2(X ^ log2(x)) = 4

    [snip]
    You require x^{\log_2(x)} = 16 (why?).

    Therefore \log_x(16) = \log_2(x). By inspection, x = 4.
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  3. #3
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    Thanks for the help, Mr Fantastic.

    It was actually a practice question for an exam. Never done anything like it before, but I guess its purpose is to extend one's thinking. Which is quite good, of course. So, since x = 4, my second method was correct for working it out?

    Thanks again!

    Nathaniel
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  4. #4
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    Quote Originally Posted by BinaryBoy View Post
    Thanks for the help, Mr Fantastic.

    It was actually a practice question for an exam. Never done anything like it before, but I guess its purpose is to extend one's thinking. Which is quite good, of course. So, since x = 4, my second method was correct for working it out?

    Thanks again!

    Nathaniel
    (\log_2(x))^2 \neq \log_2 x^2 so your working is wrong. However, it got of to a good start - better than mine in fact. Because:

    (\log_2 (x))^2 = 4 \Rightarrow \log_2(x) = \pm 2 \Rightarrow x = 4, \frac{1}{4} so there are two solutions. And  x = \frac{1}{4} is not necessarily obvious by inspection from my earlier working.
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  5. #5
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    Quote Originally Posted by BinaryBoy View Post
    Hi all,

    I've been confronted with a somewhat complex logarithm question in the form of solving for x. After a long time of thinking and trying various methods, I finally came up with two possible methods for solving the problem. However, I am not sure which (or if either) is correct, or if I have made a mistake along the way. Please clarify the correct method.

    The question is:

    log2(X ^ log2(x)) = 4

    From there, I used the power rule and brought log2(x) down the front of the equation:

    log2(x) * log2(x) = 4

    Then I added the logs together:

    log2(x) + log2(x) = 4
    The logs were multiplied, not added! You can't just change a*b to a+ b.

    log2(2x) = 4

    Then I rearranged the equation to indical form:

    2 ^ 4 = 2X

    16 = 2X

    x = 8

    Thus, x = 8. However, I want to suggest an alternate method as well that returns to this part of the equation:

    log2(x) * log2(x) = 4

    Perhaps instead of adding them together like I did in the above working, I simply multiply the two logs together to achieve:

    log2(x ^ 2) = 4

    Rearrange and solve to equal:

    2 ^ 4 = x ^ 2

    16 = x ^ 2

    x = 4


    Any help will be much appreciated!

    Thanks,

    Nathaniel
    Follow Math Help Forum on Facebook and Google+

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