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Math Help - Solving for x with logs

  1. #1
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    Solving for x with logs

    So I'm just starting to learn the properties of logs and I was doing well until I hit problems like this:


    And this: log(2x+4)+log(x-2)=1

    I know that the answer to the first is (x=1/2) and the answer to the second is (x=3) because they're practice problems with answers but I have no idea how to to get there. I've tried the steps in the book but my answers don't come out anywhere close to the correct answers. Can anyone tell me the steps to solving problems like this?

    Thanks,
    StM
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  2. #2
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    I would say \log_9(x+1) = \frac{1}{2}+\log_9x

    \log_9(x+1) = \frac{1}{2}\log_99+\log_9x

    \log_9(x+1) = \log_99^{\frac{1}{2}}+\log_9x

    \log_9(x+1) = \log_9\sqrt{9}+\log_9x

    \log_9(x+1) = \log_93+\log_9x

    \log_9(x+1) = \log_9(3\times x)

    x+1 =  3x

    You can finish from here?
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  3. #3
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    An alternate method:

    First note that x > 0 in order to not have a negative in either logarithm...


    \log_9(x + 1) = \frac{1}{2} + \log_9{(x)}

    \log_9(x + 1) - \log_9(x) = \frac{1}{2}

    \log_9{\left(\frac{x + 1}{x}\right)} = \frac{1}{2}

    \frac{x + 1}{x} = 9^{\frac{1}{2}}

    \frac{x + 1}{x} = \{-3, 3\}

    1 + \frac{1}{x} = \{-3, 3\}

    \frac{1}{x} =\{-4, 2\}

    x = \left\{-\frac{1}{4}, \frac{1}{2}\right\}.


    Now since x > 0, that means the only solution is x = \frac{1}{2}.
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  4. #4
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    Quote Originally Posted by pickslides View Post
    I would say

    \log_9(x+1) = \log_99^{\frac{1}{2}}+\log_9x

    \log_9(x+1) = \log_9\sqrt{9}+\log_9x

    \log_9(x+1) = \log_93+\log_9x

    \log_9(x+1) = \log_9(3\times x)

    x+1 =  3x

    You can finish from here?
    Yes, I can finish from x+1=3x, but I'm just slightly confused about your first step. How exactly did you get from
    \log_9(x+1) = \frac{1}{2}+\log_9x
    to
    \log_9(x+1) = \frac{1}{2}\log_99+\log_9x?

    Anyhow, I appreciate your help a lot.
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  5. #5
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    Quote Originally Posted by Prove It View Post
    An alternate method:

    First note that x > 0 in order to not have a negative in either logarithm...


    \log_9(x + 1) = \frac{1}{2} + \log_9{(x)}

    \log_9(x + 1) - \log_9(x) = \frac{1}{2}

    \log_9{\left(\frac{x + 1}{x}\right)} = \frac{1}{2}

    \frac{x + 1}{x} = 9^{\frac{1}{2}}

    \frac{x + 1}{x} = \{-3, 3\}

    1 + \frac{1}{x} = \{-3, 3\}

    \frac{1}{x} =\{-4, 2\}

    x = \left\{-\frac{1}{4}, \frac{1}{2}\right\}.


    Now since x > 0, that means the only solution is x = \frac{1}{2}.
    Thank you! This makes perfect sense!
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  6. #6
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    Use the same principle for this one:

    log(2x+4)+log(x-2)=1

    log((2x+4)(x-2)) = 1

    Expand;

    log(2x^2 +4x -4x-8) = 1

    log(2x^2-8) = 1

    Convert into indices;

    2x^2 - 8 = 10^1

    Rearrange;

    2x^2 = 18

    You can take it from here to get x = \pm 3

    Now, since, 2x + 4 > 0, x cannot be equal to -3. So, you reject the negative value.
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  7. #7
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    I think I see where i was confused now, thanks everyone
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  8. #8
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    Quote Originally Posted by SaveTheManatees View Post
    Yes, I can finish from x+1=3x, but I'm just slightly confused about your first step. How exactly did you get from
    \log_9(x+1) = \frac{1}{2}+\log_9x
    to
    \log_9(x+1) = \frac{1}{2}\log_99+\log_9x?

    Anyhow, I appreciate your help a lot.

    \log_aa =1, \forall a \in \mathbb{R}^+
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