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anonymous

  • 5 years ago

2(e^x-2)=(e^x)+7... Solve the equation

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  1. anonymous
    • 5 years ago
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    Is that \[2e^{x-2} = e^x + 7\]?

  2. anonymous
    • 5 years ago
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    ya

  3. anonymous
    • 5 years ago
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    Ok, so first off, move all the \(e^x\) terms to one side of the equal sign.

  4. anonymous
    • 5 years ago
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    how do you do that

  5. anonymous
    • 5 years ago
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    By adding or subtracting things from both sides.

  6. anonymous
    • 5 years ago
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    For example, go ahead and subtract \(e^x\) from both sides and what do you get?

  7. anonymous
    • 5 years ago
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    7=2(e^x-2)-(e^x)

  8. anonymous
    • 5 years ago
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    Good. Now lets rewrite that \(e^{x-2}\) as the product of two powers of e. Recall that \[a^b * a^c = a^{b+c}\] So what can we rewrite \[e^{x-2}\] as?

  9. anonymous
    • 5 years ago
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    e^x/e^2

  10. anonymous
    • 5 years ago
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    Correct. So that means that our equation becomes?

  11. anonymous
    • 5 years ago
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    2(e^x/e^2)-e^x=7

  12. anonymous
    • 5 years ago
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    Right. Now since both terms on the left side have an \(e^x\) factor, we can factor it out in front of the sum. a*b + a*c = a(b+c)

  13. anonymous
    • 5 years ago
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    no sure how to right it now

  14. anonymous
    • 5 years ago
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    u still there

  15. anonymous
    • 5 years ago
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    \[\frac{2}{e^2}(e^x) - e^x = 7\] \[e^x(\frac{2}{e^2} - 1)= 7\]

  16. anonymous
    • 5 years ago
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    Not sure why those fractions aren't showing up right..

  17. anonymous
    • 5 years ago
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    But anyway, then you just divide by the second bit leaving only the \(e^x\) on the left side.

  18. anonymous
    • 5 years ago
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    why don't we us ln

  19. anonymous
    • 5 years ago
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    We will, next.

  20. anonymous
    • 5 years ago
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    ok

  21. anonymous
    • 5 years ago
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    You should have \[e^x = \frac{7}{(2e^{-2}) - 1}\] Then you take the ln of both sides.

  22. anonymous
    • 5 years ago
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    so... x= ln7/(2ln -2) - ln1 not sure about denominator. Is that correct

  23. anonymous
    • 5 years ago
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    No, you have to take the ln of the whole thing, not the ln of the top divided by the ln of the bottom. If you want to break it up further you can use the property of log of a quotient is subtraction of the log. \[ln[\frac{a}{b}] = (ln\ a) - (ln\ b)\]

  24. anonymous
    • 5 years ago
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    so... x=ln7-((2ln-2)-1) ?

  25. anonymous
    • 5 years ago
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    i think i did something wrong bc u cnt have ln and then a negative. What woudit be then

  26. anonymous
    • 5 years ago
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    It should be: \[x = (ln\ 7) - (ln [2e^{-2} - 1])\] It cannot be simplified further because we can't do anything with the log of a sum. But you can plug it into your calculator if you like. I think you may have tried to do \((ln\ e^{-2})\) which isn't legal in this case, but is legal in other situations. It's not equal to \(ln -2\) though.. \[ln[e^{-2}] = -2\]

  27. anonymous
    • 5 years ago
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    thanks for everything

  28. anonymous
    • 5 years ago
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    Of course!

  29. anonymous
    • 5 years ago
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    it said not real in calculator

  30. anonymous
    • 5 years ago
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    Hrmph.. That's true actually. \(2e^{-2}\) is less than 1, so when you subtract 1 from it you'll have a negative number and you cannot take the ln of a negative.

  31. anonymous
    • 5 years ago
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    We must have missed something

  32. anonymous
    • 5 years ago
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    Nope, that's the correct solution to this equation. It has no real soltions.

  33. anonymous
    • 5 years ago
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    Solutions rather. Double check what I wrote originally.

  34. anonymous
    • 5 years ago
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    what u originaly wrote is corect

  35. anonymous
    • 5 years ago
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    must be no solutions thanks again

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