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    Solve to find x in terms of a where a>0 and a isnt 100

    a^x=10^2x+1 the 2x+1 is all raised. I'm not reall too sure how to start the question i know the rules of logs just don't know how to apply it here.
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    Take the logarithm to base 10 of both sides (or any base, but base 10 simplifies it a bit). Then use the rule that \alpha \log k = \log (k^{\alpha}) and rearrange to solve for x.
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    Take logs. What've you got?
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    Take logs of each side, expand the bracket on the right, and then it's just moving stuff around.
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    (Original post by nuodai)
    Take the logarithm to base 10 of both sides (or any base, but base 10 simplifies it a bit). Then use the rule that \alpha \log k = \log (k^{\alpha}) and rearrange to solve for x.
    is this right log10a^x=log1010^2x+1

    then does it become xloga=2x+1 log1 not sure if this bit is correct
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    Nasty question lol...

    did you get it down to... "x/(2x+1) = 1/loga" ?
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    (Original post by hazbaz)
    is this right log10a^x=log1010^2x+1

    then does it become xloga=2x+1 log1 not sure if this bit is correct
    Not quite, no; the RHS should be (2x+1)\log_{10} 10. Or, better, just use the fact that \log_b b^a = a (by the definition of a logarithm).
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    (Original post by nuodai)
    Not quite, no; the RHS should be (2x+1)\log_{10} 10. Or, better, just use the fact that \log_b b^a = a (by the definition of a logarithm).
    Therefore xloga = 2x+1 ?? lol im confused aswell
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    (Original post by Extricated)
    Therefore xloga = 2x+1 ?? lol im confused aswell
    Yup.
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    (Original post by nuodai)
    Yup.
    Sorry, this may sound like a stupid question (tired), but whats the next step
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    (Original post by Extricated)
    Sorry, this may sound like a stupid question (tired), but whats the next step
    How would you solve the equation px=qx+r for x (where p,q,r are constants)? This is no different.
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    (Original post by nuodai)
    How would you solve the equation px=qx+r for x (where p,q,r are constants)? This is no different.
    px = qx + r

    px-qx = r

    x(p-q) = r

    x = r/(p-q)


    so in this case :

    xloga = 2x+1

    xloga - 2x = 1

    x(loga-2) = 1

    X = 1/loga -2
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    (Original post by Extricated)
    px = qx + r

    px-qx = r

    x(p-q) = r

    x = r/(p-q)


    so in this case :

    xloga = 2x+1

    xloga - 2x = 1

    x(loga-2) = 1

    x = 1/loga-2

    is A in this case 10? maybe thats where im going wrong..wow i feel so dumb
    I don't know how much I want to say because you're kind of hijacking the OP's thread here, but basically... you're not solving for a. The constant a is fixed; instead you're trying to find x in terms of a... which is what you just did. All we're told is that a>0 and a \ne 100, so a could be 10 or it could be 2 or 3 or even 410\pi.
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    (Original post by nuodai)
    I don't know how much I want to say because you're kind of hijacking the OP's thread here, but basically... you're not solving for a. The constant a is fixed; instead you're trying to find x in terms of a... which is what you just did. All we're told is that a>0 and a \ne 100, so a could be 10 or it could be 2 or 3 or even 410\pi.
    Cool.
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    i did it anyway :P thanks i'm just not that good n logarthmic equations going to practice them lat on
 
 
 
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