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Math Help - solving logarithmic equations

  1. #1
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    solving logarithmic equations

    I need help solving for x. i am pretty new at this, and not too sure what the first couple steps would be.

    log6 X + log6 (X-5)=2
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by extraordinarymachine View Post
    I need help solving for x. i am pretty new at this, and not too sure what the first couple steps would be.

    log_6 X + log6 (X-5)=2
    remember your properties

    Identityloga(xy)=log_(a)x + log_(a)y


    Example(a) log_(2)16 = log_(2)8 + log_(2)2

    so \log_{6}x(x-5)=2

    by the inverse property

    6^2=x(x-5)

    peace out.
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  3. #3
    Senior Member I-Think's Avatar
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    Convert the R.H.S. to logarithmic form in base 6. (c=b^a\rightarrow a=log_{b}c)
    Use addition formula. (log_{b}K+log_{b}L=log_b{KL})
    Drop logs.
    Have fun solving, because it's all about having fun.

    In action.

    log6 X + log6 (X-5)=2
    log6 X + log6 (X-5)=log6 36
    log6 (X(X-5))=log6 36
    X^2-5X=36

    Have fun solving.
    Last edited by I-Think; June 9th 2009 at 02:08 PM.
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  4. #4
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    Quote Originally Posted by I-Think View Post
    Convert the R.H.S. to logarithmic form in base 6. (c=b^a\rightarrow a=log_{b}c)
    Use addition formula. (log_{b}K+log_{b}L=log_b{KL}
    Drop logs.
    Have fun solving, because it's all about having fun.

    In action.

    log6 X + log6 (X-5)=2
    log6 X + log6 (X-5)=log6 36
    log6 (X(X-5))=log6 36
    X^2-5X=36

    Have fun solving.
    @OP: Note that only solutions such that x > 5 are valid.
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  5. #5
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    Thanks for all your help everyone. i think i understand now. at least to do these types of questions ha ha.
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