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anonymous

  • 5 years ago

solve the following for x: 3e^(-0.15x+1)=6

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  1. anonymous
    • 5 years ago
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    First isolate the \(e^{-.15x + 1}\)

  2. anonymous
    • 5 years ago
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    So divide 3 out?

  3. anonymous
    • 5 years ago
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    Yes

  4. anonymous
    • 5 years ago
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    leaves you with e^(-0.15x+1)=(6/3) and I'm still stuck...

  5. anonymous
    • 5 years ago
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    Well 6/3 = 2. Then you take the natural log of both sides

  6. anonymous
    • 5 years ago
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    Remember that \[ln\ e^a = a\]

  7. anonymous
    • 5 years ago
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    obviously correct me if I'm wrong but would that leave me with -0.15x+1/x-2/x=0?

  8. anonymous
    • 5 years ago
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    I don't think so. \[3e^{(-0.15x+1)}=6\] \[\implies e^{-.15x + 1} = 2\] \[\implies ln\ e^{-.15x + 1} = ln 2\] \[\implies -.15x + 1 = ln 2\] \[\implies -.15x = (ln2)-1\] \[\implies x = \frac{(ln2) -1}{-.15}\]

  9. anonymous
    • 5 years ago
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    Oh. Hmm what if you have a polynomial. The example I have is 2^(2x)+2^(x)-6=0. Are the steps much different?

  10. anonymous
    • 5 years ago
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    Yes

  11. anonymous
    • 5 years ago
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    For that you'd want to start by solving the quadratic.

  12. anonymous
    • 5 years ago
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    Let \(k = 2^x \implies 2^{2x} = (2^x)^2 = k^2\) \[\implies 2^{2x} + 2^x -6 = k^2 + k -6 = 0\] Use the quadratic formula (or factoring) to solve for k.

  13. anonymous
    • 5 years ago
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    0=-3, 2?

  14. anonymous
    • 5 years ago
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    k = -3 and k = 2.

  15. anonymous
    • 5 years ago
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    whoops that's what I meant!

  16. anonymous
    • 5 years ago
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    Since k = \(2^x\) that means that \(2^x = -3\) and \(2^x = 2\) are solutions.

  17. anonymous
    • 5 years ago
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    So to find x we again take the log of both sides.

  18. anonymous
    • 5 years ago
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    from which equation?

  19. anonymous
    • 5 years ago
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    From those last two equations.

  20. anonymous
    • 5 years ago
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    \[2^x = -3\] and \[2^x = 2\]

  21. anonymous
    • 5 years ago
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    Sorry for the extremely late reply wouldn't that lead you with ln2^x=-3ln and ln2^x=2ln? but to get x alone I'm lost

  22. anonymous
    • 5 years ago
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    Ah, well when you take the log of something you can bring out (and down) the exponent. \[ln\ 2^x= x(ln\ 2)\]

  23. anonymous
    • 5 years ago
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    So you have \[x(ln\ 2) = ln\ 2\] and \[x(ln\ 2) = ln\ -3\] But (ln -3) is not defined, so the first solution is the only valid one. And when you divide both sides by (ln 2) you have x = 1.

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