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Math Help - Logarithm Function

  1. #1
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    Logarithm Function

    Can someone give me a hand?

    The problem is Log3 (3x-2)=3

    I have A= 3

    and then I have 3^3 (3x-2)

    X=?

    I can't figure out how to sove it. I know it is a fraction but can anyone tell me what it is or how to solve it?
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by dimanre View Post
    Can someone give me a hand?

    The problem is Log3 (3x-2)=3

    I have A= 3

    and then I have 3^3 (3x-2)

    X=?

    I can't figure out how to sove it. I know it is a fraction but can anyone tell me what it is or how to solve it?
    \begin{gathered}<br />
  {\log _3}\left( {3x - 2} \right) = 3; \hfill \\<br />
  3x - 2 = {3^3}; \hfill \\<br />
  3x - 2 = 27. \hfill \\ <br />
\end{gathered}

    You still have to solve linear equation 3x - 2 = 27.
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  3. #3
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    Ok ..so (3x-2) =27

    ok..so 3x-2=27

    What is the easiest way to figure it out? obviously 3 * 10 =30 -2 would be 29...so that is not the answer. I know this may sound dumb but how do i then solve for x? I am very new to calculus. I don't need you to give me the answer if you rather not..but can you direct me a little further? thank you..
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  4. #4
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    Is my answer correct

    if my math is correct..i get 29..is that correct?
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  5. #5
    Senior Member DeMath's Avatar
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    Quote Originally Posted by dimanre View Post
    ok..so 3x-2=27

    What is the easiest way to figure it out? obviously 3 * 10 =30 -2 would be 29...so that is not the answer. I know this may sound dumb but how do i then solve for x? I am very new to calculus. I don't need you to give me the answer if you rather not..but can you direct me a little further? thank you..
    \begin{gathered}3x - 2 = 27; \hfill \\<br />
  3x = 29; \hfill \\<br />
  x = \frac{{29}}<br />
{3}. \hfill \\ \end{gathered}
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  6. #6
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    Thank you very much

    Just as you gave me the answer I found an example in my text. But I did it the other way and got 3/29 not 29/3. Thank you. I know i have a long way to go to learn..but I am honestly trying hard. THanks and I am sure i will be having more questions.
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