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Math Help - Exponential and Logarithmic Equations

  1. #1
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    Exponential and Logarithmic Equations

    Hey guys,


    Solve the equation. Express irrational solutions in exact form and as a decimal rounded to three decimal places.
    You don't have to do the decimal part.

    the answer is log (0.2) / (log3 - log0.2)

    0.2^X+1 = 3^X

    what I did is

    x+1Log 0.2 = xLog3

    and then i divided Log 0.2 from both sides but that doesn't work.
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  2. #2
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    Re: Exponential and Logarithmic Equations

    Use brackets when your indexes are more than 1 letter.
    You correctly took logs of both sides. Your equation (x+1)Log 0.2 = xLog3 is just like (x+1)A = xB, find x from that like you normally would.
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  3. #3
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    Re: Exponential and Logarithmic Equations

    Quote Originally Posted by Shakarri View Post
    Use brackets when your indexes are more than 1 letter.
    You correctly took logs of both sides. Your equation (x+1)Log 0.2 = xLog3 is just like (x+1)A = xB, find x from that like you normally would.

    okay, so I divide log 3 on both sides which gets me x+1Log0.2 / Log3


    but the answer is (0.2) / (log3 - log0.2)

    what do I do with the x + 1 ?
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  4. #4
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    Re: Exponential and Logarithmic Equations

    \frac{(x+1)Log0.2}{Log3}=x

    x\frac{Log0.2}{Log3}+\frac{Log0.2}{Log3}=x

    Solve for x
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  5. #5
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    Re: Exponential and Logarithmic Equations

    Oh man, sorry, but I'm completely lost. I don't know how you got two terms. Also, I have no idea how to solve for x right there.
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  6. #6
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    Re: Exponential and Logarithmic Equations

    Log of a number is a constant, its just like any other number you'd get in an equation.
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  7. #7
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    Re: Exponential and Logarithmic Equations

    Ohhhhhh so you just distributed the (x+1) with Log0.2 I see.
    ahh, I see. So, now I move Log0.2 / Log3 to the right of the equation? Now, I have xlog0.2/log3 = - log0.2/log3

    I don't think that's correct :\
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  8. #8
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    Re: Exponential and Logarithmic Equations

    You were told in Shakarri's first response that this is the same as A(x+ 1)= Bx (with A= log(0.2) and B= log(3)). That is the same as Ax+ A= Bx so, subtracting Ax from both sides A= (B- A)x. Can you solve that?
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  9. #9
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    Re: Exponential and Logarithmic Equations

    Yes, I got log0.2 = (log3 - log0.2)x / log0.2

    it's just reversed. log0.2 has to be on top.
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