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Math Help - Logarithms problems

  1. #1
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    Question Logarithms problems

    hey i just have no clue in how to attempt this,
    anyone shine a light?
    Q: solve equations
    a)logx+3 = 4
    b) log(x+3) = 4
    C) logx+log3 = 4

    Q: Log(x+1) - log(x-6) +2logx i know here you have apply the rules of log, but i have not really been used to dealing with brackets in these situations
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by JorCherryhead View Post
    hey i just have no clue in how to attempt this,
    anyone shine a light?
    Q: solve equations
    a)logx+3 = 4
    b) log(x+3) = 4
    C) logx+log3 = 4

    Q: Log(x+1) - log(x-6) +2logx i know here you have apply the rules of log, but i have not really been used to dealing with brackets in these situations
    Hi JorCherryhead,

    Looks like you need a text book or a good tutorial. Try this one: College Algebra Tutorial on Logarithmic Equations

    I'll do the first one, and get you started on the other two.

    a) \log x+3=4

    \log x = 1

    x=10^1=10

    b) \log (x+3)=4

    x+3=10^4

    c) \log x+\log 3 =4

    \log 3x=4

    3x=10^4
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  3. #3
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    thankyou=] ill have a look at the help guide.
    much appreciated
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  4. #4
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    Exclamation

    hey with this equation what would i do first
    or does it not matter if i work with the product rule then the quotient rule or vis versa, i=or is one way easier.
    as with the examples on the website you showed me, when they were working there equations out, they seemed to be left with a exponent on the other side of the equals sign, and unforunatly i have not got that on this equation
    log(x+1) - log(x-6) +2logx
    at first i did
    log (x+1/x-6) +log x^2
    and im not sure but is there a way to make the power (2) to become the exponent or am i talking nonsence, as if i could i would do
    somehow
    log(X+1/X-6)=2
    then i could
    10^2 = x+1/x-6
    and then from that solve x
    but im not sure
    ?
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  5. #5
    A riddle wrapped in an enigma
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    Quote Originally Posted by JorCherryhead View Post
    hey with this equation what would i do first
    or does it not matter if i work with the product rule then the quotient rule or vis versa, i=or is one way easier.
    as with the examples on the website you showed me, when they were working there equations out, they seemed to be left with a exponent on the other side of the equals sign, and unforunatly i have not got that on this equation
    log(x+1) - log(x-6) +2logx
    at first i did
    log (x+1/x-6) +log x^2
    and im not sure but is there a way to make the power (2) to become the exponent or am i talking nonsence, as if i could i would do
    somehow
    log(X+1/X-6)=2
    then i could
    10^2 = x+1/x-6
    and then from that solve x
    but im not sure
    ?
    First of all, \log(x+1) - \log (x-6) +2 \log x is not an equation.

    Did you miscopy it? There's no '=' sign.
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  6. #6
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    =] that may be why what i was doing wasn't making alot of sence.
    so would it still be
    log(X+1/X-6) +logx^2
    then
    logx^3+x^2 divided by x-6
    ?
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