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Math Help - Basic Logirithms & Basic expanding the brackets

  1. #1
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    Basic Logirithms & Basic expanding the brackets

    Hello everyone,

    I need help with logarithms that are in fraction form.

    such as
    evaluate the following for x:


    log ^ 2 x = 1/7

    log ^ 2 x = 1/5

    log ^ 2 x = 2/7


    I also need to be sure if expanding this equation for e.g is correct?

    expand the following:

    (a + b) (a + b + c)

    a^2 + ab + ac + ba + b^2 + bc

    & do I plus the powers or times them in the brackets?


    expand the following:


    (a^2 + b^2) (a^3 + b^3 + c^2)

    a^5 + a^2b^5 + a^2c^2 + b^2a^3 + b ^ 5 + b ^2c^ 2
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  2. #2
    Super Member Bacterius's Avatar
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    evaluate the following for x
    You must mean "solve the following for x" ? If yes, then :

    log(2x) = \frac{\ 1}{7}
    2x = 10^{\frac{\ 1}{7}}
    x = \frac{\ 10^{\frac{\ 1}{7}}}{2}

    (a + b)(a + b + c)
    a(a + b + c) + b(a + b + c)
    a^2 + ab + ac + ba + b^2 + bc
    a^2 + b^2 + 2ab + ac + bc

    But maybe I don't understand what you wrote, try using the LaTeX editor ?
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Detroit, MI
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    Quote Originally Posted by student0451 View Post
    Hello everyone,

    I need help with logarithms that are in fraction form.

    such as
    evaluate the following for x:


    log ^ 2 x = 1/7

    log ^ 2 x = 1/5

    log ^ 2 x = 2/7


    I also need to be sure if expanding this equation for e.g is correct?

    expand the following:

    (a + b) (a + b + c)

    a^2 + ab + ac + ba + b^2 + bc

    & do I plus the powers or times them in the brackets?


    expand the following:


    (a^2 + b^2) (a^3 + b^3 + c^2)

    a^5 + a^2b^5 + a^2c^2 + b^2a^3 + b ^ 5 + b ^2c^ 2
    1. Do you mean

    \log_2x=\frac{1}{7} or \log^2x=\frac{1}{7} ?


    2. Collect your like terms
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  4. #4
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    i am so sorry about my typing,

    http://www.mathhelpforum.com/math-he...6f55baf4-1.gif

    it is the base of 2

    log 2 x = 1/7

    so 2 is the base of the log


    and thank you for helping me.
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  5. #5
    Super Member Bacterius's Avatar
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    Wellington
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    Reply

    Then it becomes (I think) :

    log_2(x) = \frac{\ 1}{7}
    x = 2^{\frac{\ 1}{7}}

    When writing indicators, use the underscore char, it is much more appropriate : log_2(x).
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Detroit, MI
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    Quote Originally Posted by student0451 View Post
    i am so sorry about my typing,

    http://www.mathhelpforum.com/math-he...6f55baf4-1.gif

    it is the base of 2

    log 2 x = 1/7

    so 2 is the base of the log


    and thank you for helping me.
    \log_2x=\frac{1}{7}\Rightarrow{x}=2^{1/7}=\sqrt[7]{2}
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  7. #7
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    yep

    and thank you, vondera19 & Bacterius.

    I knew there was something wrong between the way I expanded brackets.
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