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Math Help - Exponential and Logarithmic Equations

  1. #1
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    Exponential and Logarithmic Equations

    Hey guys,

    log(x+3) - logx = 1


    so I got 1 = (x+3) / (x)

    and the book says the answer is 1/3

    I don't know how it got this

    any help? Thanks
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  2. #2
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    Re: Exponential and Logarithmic Equations

    Is that the common logarithm? Base 10? If so then you have log(x+3)- log(x)= log\left(\frac{x+ 3}{x}\right)= 1 and then "get rid of" the logarithm by taking 10 to the power of each side. Instead of getting \frac{x+ 3}{x}= 1 you should have \frac{x+ 3}{x}= 10^1= 10.

    Can you solve \frac{x+ 3}{x}= 10?
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  3. #3
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    Re: Exponential and Logarithmic Equations

    Quote Originally Posted by Cake View Post
    log(x+3) - logx = 1
    and the book says the answer is 1/3 I don't know how it got this
    .


    The answer depends upon the base.

    If {\log _b}(x + 3) - {\log _b}(x) = 1 then {\log _b}\left( {\frac{{x + 3}}{x}} \right) = 1.

    Then \frac{x+3}{x}=b.

    So I don't know where that answer comes from.
    Thanks from Cake
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  4. #4
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    Re: Exponential and Logarithmic Equations

    Correct, it is base 10. Ahh I see. So whenever I have " = 1 " and with a base of 10. It automatically equals to 10^1?

    and yes, thank you. That gives me 1 / 3. Thanks a lot for clearing this up. I would have never gotten this.
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