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Math Help - another log problem

  1. #1
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    another log problem

    Hi there, I have yet another log problem that's got me tearing my hair out. e^(2x)-e^(x+3)-e^(x+1)+e^4=0. I tried taking e^4 to the RHS and taking the ln of both sides and got 0=0...wtf.
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  2. #2
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    Quote Originally Posted by flashylightsmeow View Post
    Hi there, I have yet another log problem that's got me tearing my hair out. e^(2x)-e^(x+3)-e^(x+1)+e^4=0. I tried taking e^4 to the RHS and taking the ln of both sides and got 0=0...wtf.
    I can't think of how to solve it either, but you did make an error. ln(a + b) is not equal to ln(a) + ln(b). So ln[e^(2x)-e^(x+3)-e^(x+1)] is not (2x) - (x + 3) - (x + 1)

    -Dan
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  3. #3
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    <br />
e^{2x}-(e^3+e)e^x+e^4=0<br />

    <br />
y=e^x<br />

    it is quadratic equation.
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  4. #4
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    Quote Originally Posted by flashylightsmeow View Post
    e^(2x)-e^(x+3)-e^(x+1)+e^4=0. I tried taking e^4 to the RHS and taking the ln of both sides and got 0=0...wtf.
    I have solved this. x=3~\&~x=1.
    It is tricky. Notice that it can be written as:
    \displaystyle e^{2x}-(e^3+e^2)e^x+e^4=0
    That is a quadric in \displaystyle e^x.
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  5. #5
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    Hi there, firstly thank you topsquark, zzzoak and plato for looking at this. I have 2 queries, Plato your quadratic has (e^3+e^2)e^x but wouldn't that give e^(x+3)-e^(x+2)? And how do you go from this quadratic equation to the solution. I just can't see it. I feel like throwing my computer out of the window!
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  6. #6
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    Wait, I think I get it, do you then factorise to get (e^x - e^3)(e^x - e^1)=0 ?
    Also how are you getting the formula to display properly?
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  7. #7
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    Quote Originally Posted by flashylightsmeow View Post
    Hi there, firstly thank you topsquark, zzzoak and plato for looking at this. I have 2 queries, Plato your quadratic has (e^3+e^2)e^x but wouldn't that give e^(x+3)-e^(x+2)?
    No, it wouldn't. How did that "+" in (e^3+ e^2)(e^x) become a "-"?

    Wait, I think I get it, do you then factorise to get (e^x - e^3)(e^x - e^1)=0 ?
    Yes, good. e^3 and e add to give e^3+ e and multiply to give e^4 so that factorization is correct. And, now the solution to the problem is easy.

    Also how are you getting the formula to display properly?
    Use LaTex. On this board you each LaTex formula begins with [ math ] and ends with [ /math ], without the spaces which I had to put in so they would show. (Unfortunately, that part varies from website to website that uses LaTex.) For example [ math ] e^{x+ y}[ /math ] without the spaces is e^{x+y}.

    You can double click on any formula to see the code used. Also, there is a "Latex tutorial" area under the main menu.
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  8. #8
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    Hello, flashylightsmeow!

    \text{Solve for }x\!:\;\;e^{2x}-e^{x+3}-e^{x+1}+e^4\:=\:0

    We have: . e^{2x} - e^3\!\cdot\!e^x - e\!\cdot\!e^x + e^4 \;=\;0

    . . . . . . . . e^{2x} - (e^3+e)e^x + e^4 \;=\;0

    Factor:. . . . . (e^x - e)(e^x - e^3) \;=\;0


    Got it?

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  9. #9
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    You guys have saved me from putting my head through a wall. Many thanks. I was able to solve a similar problem by myself, now I have to apply this in financial contexts...ugh.
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