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Math Help - Logarithms

  1. #1
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    Logarithms

    Not sure how to put this in here, so hopefully i can do my best
    Please help with this

    log3(logx(log416))=-1
    The 3, x and 4 are all the base, so they should be lower than the log and 16. I hope that i said and did that right
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  2. #2
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    Quote Originally Posted by nuckers View Post
    Not sure how to put this in here, so hopefully i can do my best
    Please help with this

    log3(logx(log416))=-1
    The 3, x and 4 are all the base, so they should be lower than the log and 16. I hope that i said and did that right
    Use underscore, which should be SHIFT plus the key right next to "0" and slightly above and to the right of "P" if your keyboard is like mine.

    log_3(log_x(log_416))=-1

    log_x(log_416)=3^{-1}=\frac{1}{3}

    log_416=x^{\frac{1}{3}}

    2=x^{\frac{1}{3}}

    x=2^3 = 8
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  3. #3
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    Quote Originally Posted by undefined View Post
    Use underscore, which should be SHIFT plus the key right next to "0" and slightly above and to the right of "P" if your keyboard is like mine.

    log_3(log_x(log_416))=-1

    log_x(log_416)=3^{-1}=\frac{1}{3}

    log_416=x^{\frac{1}{3}}

    2=x^{\frac{1}{3}}

    x=2^3 = 8
    wow, that was fast and easy, i couldn't remember how the bases work, but i do know, boot the log, thanks so much for your help.
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    so for this one, i got an answer of x=5, is this correct

    _62log_6x+log_6x=125
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  5. #5
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    Quote Originally Posted by nuckers View Post
    so for this one, i got an answer of x=5, is this correct

    _62log_6x+log_6x=125
    Something strange going on with the 6 and 2 at the beginning. Could you clarify what this is supposed to be?
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  6. #6
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    Quote Originally Posted by undefined View Post
    Something strange going on with the 6 and 2 at the beginning. Could you clarify what this is supposed to be?
    the 2log6x+log6x is all raised, not sure how else to explain it
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  7. #7
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    Quote Originally Posted by nuckers View Post
    the 2log6x+log6x is all raised, not sure how else to explain it
    Is it like this?

    \displaystyle{6^{2log_6x+log_6x}=125}

    (I know it's kind of hard to read, but if it's right you should be able to tell.)
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  8. #8
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    Quote Originally Posted by undefined View Post
    Is it like this?

    \displaystyle{6^{2log_6x+log_6x}=125}

    (I know it's kind of hard to read, but if it's right you should be able to tell.)
    Thats exactly how it's written, good job
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  9. #9
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    Quote Originally Posted by nuckers View Post
    Thats exactly how it's written, good job
    \displaystyle{6^{2\log_6x+\log_6x}=125}

    2\log_6x+\log_6x=\log_6(125)

    3\log_6x=\log_6(125)

    \log_6(x^3)=\log_6(125)

    x^3=125

    x=5

    So, to answer your question, yes.
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  10. #10
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    Quote Originally Posted by undefined View Post
    \displaystyle{6^{2\log_6x+\log_6x}=125}

    2\log_6x+\log_6x=\log_6(125)

    3\log_6x=\log_6(125)

    \log_6(x^3)=\log_6(125)

    x^3=125

    x=5

    So, to answer your question, yes.
    awesome, exactly what i got, just wanted to double check, some of these logs are so confusing, thanks alot
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