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Math Help - Trouble with X intercept question

  1. #1
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    Trouble with X intercept question

    Hi, I'm having some issues with this problem:

    f(x) = -2e^{5x}+6
    I need to solve for the x intercept of the graph of f.

    I tried to do it like so:

    0 = -2e^{5x}+6
    2e^{5x} = 6
    \log(2e^{5x}) = \log(6)

    5x\log(2e) = \log(6)

    5x = \frac{\log(6)}{\log(2e)}

    x = \frac{\frac{\log(6)}{\log(2e)}}{5}

    That gives me: .2116484012

    Using my graphing calculator to get the intersection of:
    y1 = -2e^{5x}+6
    y2 = 0

    Gives an answer of: .21972246


    Would someone mind explaining what I've done wrong? Thanks!
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  2. #2
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    e^(i*pi)'s Avatar
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    The x intercept is where the graph crosses the x axis.

    In other words you need to find f(0)
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  3. #3
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    Isn't f(0) = 4 ? Or, the y intercept?
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  4. #4
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    Quote Originally Posted by cb220 View Post
    Isn't f(0) = 4 ? Or, the y intercept?
    Yes f(0) = 4 which is the/an x-intercept which is what the question asked for

    The y intercept is

    2e^{5x} = 6

    e^{5x} = 3

    5x = \ln(3)

    x = \dfrac{1}{5} \ln(3) \approx 0.220


    -------------

    You went wrong between these two steps: \log(2e^{5x}) = \log(6) and

    5x\log(2e) = \log(6)

    You can't do that because the exponent is only on e. For that to work you'd need the exponent on the 2 and the e. Instead you'd have to split the logs into \log(2) + 5x \log(e) = \log(6)

    5x\log(e) = \log(6) - \log(2) = \log(3)

    x = \dfrac{\log(3)}{5\log(e)}  \approx 0.220


    edit 2: You can use the change of base rule to see that \ln(3) = \dfrac{\log(3)}{\log(e)}
    Last edited by e^(i*pi); October 2nd 2010 at 12:44 PM. Reason: white space
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  5. #5
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    Ah, I see how I went wrong with my exponent not being attached to both terms. Thanks!

    edit: However I'm still a bit confused on the x vs y intercepts. Putting my answers into the webwork form and using 4 as the y intercept and 2.1972246 as the x intercept came back as correct. I thought f(0) is the y intercept because the y intercept is where x=0 ? Conversely the x intercept would be where y = 0. Am I missing/forgetting something with that?
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  6. #6
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    Quote Originally Posted by cb220 View Post
    Ah, I see how I went wrong with my exponent not being attached to both terms. Thanks!

    edit: However I'm still a bit confused on the x vs y intercepts. Putting my answers into the webwork form and using 4 as the y intercept and 2.1972246 as the x intercept came back as correct. I thought f(0) is the y intercept because the y intercept is where x=0 ? Conversely the x intercept would be where y = 0. Am I missing/forgetting something with that?
    From what I gather you have the definition of intercept the wrong way around.

    The x intercept is where the graph crosses or touches the x axis. Thus, by deifintion the x value of that point(s) must always be 0. On your graph this makes (0,4) a point.

    The y intercept is where it crosses or touches the y axis and hence y=0 and you will have (x,0) which usually requires more effort to find.


    To find the x-intercept you calculate f(0) and to find the y intercept you find f(x)=0. I hope that's clearer
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  7. #7
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    Quote Originally Posted by e^(i*pi) View Post
    The x intercept is where the graph crosses or touches the x axis. Thus, by deifintion the x value of that point(s) must always be 0. On your graph this makes (0,4) a point.
    Isn't (0,4) a point on the y axis? 0 over and 4 up?
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  8. #8
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    Yes, you're right. Turns out I got the intercepts the wrong way round

    4 would be the y intercept and 2.20 would be the x intercept
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