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Math Help - Tangent to e^x through origin

  1. #1
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    Tangent to e^x through origin

    Find the value of A such as that the tangent to y=e^x is a line through origin when x=a

    This is what I've done so far,

    Derivative of e^x is e^x, so the slope is e^x

    (y-e^a)=e^a(x-a)
    0-e^a=e^a(0-a)
    -e^a=-a(e^a)

    That's what I get in the end, how can I solve for a? Did I make an error? This corresponds to a question asking a similar question for lnx, where a would be the value e, so I suppose there has to be some connection....
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Here is what you do

    Quote Originally Posted by theowne View Post
    Find the value of A such as that the tangent to y=e^x is a line through origin when x=a

    This is what I've done so far,

    Derivative of e^x is e^x, so the slope is e^x

    (y-e^a)=e^a(x-a)
    0-e^a=e^a(0-a)
    -e^a=-a(e^a)

    That's what I get in the end, how can I solve for a? Did I make an error? This corresponds to a question asking a similar question for lnx, where a would be the value e, so I suppose there has to be some connection....
    -e^{a}=-a\cdot{e^{a}}... a=1?...o you made a mistake I think you should have y-0=e^{a}(x-0)...so y=e^{a}x....a=0...you just forgot that the slope for tangent line must be the same point as is shown in the equation for a tangent line y-f(x_1)=f'(x_1)(x-x_1)
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  3. #3
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    That's what I figured, but what connection exists between that and the value e? The question seems to imply a relation between the answer here, and e..
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    The seeming connection would be

    Quote Originally Posted by theowne View Post
    That's what I figured, but what connection exists between that and the value e? The question seems to imply a relation between the answer here, and e..
    f(x)=\ln(x) both give the equation y=x? since y-e=\ln(e)(x-e)...so y=\ln(e)x-\ln(e)e+e=x...andy maybe that \ln(e)=1,e^{0}=1?
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  5. #5
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    I think my earlier form was correct....(which results in a=1)..I plugged it into graphmatica and I can see the relationship more clearly now...for ln(x), it's when x=e, for e^x, it's when y=e. Thanks.....
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Hmm

    Quote Originally Posted by theowne View Post
    I think my earlier form was correct....(which results in a=1)..I plugged it into graphmatica and I can see the relationship more clearly now...for ln(x), it's when x=e, for e^x, it's when y=e. Thanks.....
    Maybe I am misunderstanding your question then...but based on you think what you just said...maybe you are supposed to realize that since \ln(x) and e^{x} are inverse functions their domain and ranges are switched
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