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anonymous

  • one year ago

Solve for x using the natural logarithm:

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  1. jdoe0001
    • one year ago
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    kinda hard to do, when the expression is invisible

  2. anonymous
    • one year ago
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    \[2\times4^{x} = 9.5e ^{-2x}\]

  3. anonymous
    • one year ago
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    Thank you for your patience, @jdoe0001

  4. anonymous
    • one year ago
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    @SithsAndGiggles Can you help?

  5. jdoe0001
    • one year ago
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    hmmm can you pot a quick screenshot of the material?

  6. anonymous
    • one year ago
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    @zepdrix Can you help? I'm desperate :(

  7. zepdrix
    • one year ago
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    :D

  8. zepdrix
    • one year ago
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    Oh you left :c

  9. anonymous
    • one year ago
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    I'm back!

  10. zepdrix
    • one year ago
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    :D

  11. anonymous
    • one year ago
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    :D

  12. zepdrix
    • one year ago
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    \[\large\rm 2\cdot4^{x} = 9.5\cdot e ^{-2x}\]I don't like decimals, fractions are better, so I'm gonna do something a little sneaky. Let's umm.... let's multiply both sides by 2, then divide each side by 4,\[\large\rm 4^x=\frac{19}{4}e^{-2x}\]

  13. zepdrix
    • one year ago
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    Then take natural log each side I guess, ya? :)

  14. zepdrix
    • one year ago
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    \[\large\rm \ln\left(4^x\right)=\ln\left(\frac{19}{4}e^{-2x}\right)\]We have to apply a bunch of fun log rules to solve for x.

  15. zepdrix
    • one year ago
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    \[\large\rm \color{orangered}{\log(a^b)=b\cdot \log(a)}\]Do you see how this orange rule might help us on the left side of our equation?

  16. anonymous
    • one year ago
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    Ok!

  17. anonymous
    • one year ago
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    Give me one sec

  18. anonymous
    • one year ago
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    Wait why doesn't 4 cancel out when you divide it?

  19. zepdrix
    • one year ago
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    Umm if that was too confusing, we can just do it this way instead :)\[\large\rm 2\cdot4^{x} = 9.5\cdot e ^{-2x}\]Dividing by 2 gives us:\[\large\rm 4^{x} = 4.75\cdot e ^{-2x}\]I just didn't like the decimal value, so I changed it to a fraction, but we can leave it like this if it's easier to understand.

  20. anonymous
    • one year ago
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    Ok

  21. zepdrix
    • one year ago
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    So taking log of each side gives us\[\large\rm \ln\left(4^x\right)=\ln\left(4.75\cdot e^{-2x}\right)\]

  22. anonymous
    • one year ago
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    So you get x(ln4) = ln(4.75e^-2x) ?

  23. anonymous
    • one year ago
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    And now ln and e cancel each other out

  24. zepdrix
    • one year ago
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    Left side looks good. Woops! Don't try to cancel out the e just yet! Gotta do some more simplifying before you can do that.

  25. zepdrix
    • one year ago
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    \[\large\rm \color{orangered}{x\ln4}=\ln\left(4.75\cdot e^{-2x}\right)\]How bout the right side? How can we apply this blue rule?\[\large\rm \color{royalblue}{\log(a\cdot b)=\log(a)+\log(b)}\]

  26. anonymous
    • one year ago
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    x(ln4)= ln(4.75) + ln(e^-2x) ?

  27. zepdrix
    • one year ago
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    Ok great! Now we don't have the number in front of the e, so we can "cancel" them out like you wanted.

  28. anonymous
    • one year ago
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    xln4= ln4.75 - 2x ?

  29. zepdrix
    • one year ago
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    \[\large\rm x \ln4=\ln4.75+\ln e^{-2x}\]\[\large\rm x \ln4=\ln4.75-2x\]mmmm k good!

  30. zepdrix
    • one year ago
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    Let's get all of our x's to one side and try some factoring.

  31. anonymous
    • one year ago
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    x(ln4) + 2x= ln4.75 ?

  32. zepdrix
    • one year ago
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    cool :) the terms on the left side both have something in common. try to factor it out :O

  33. anonymous
    • one year ago
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    x(ln4 + 2)= ln4.75 ?

  34. zepdrix
    • one year ago
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    Good good good, how you gonna wrap it up? :)

  35. anonymous
    • one year ago
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    x = 0.4601 ?

  36. zepdrix
    • one year ago
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    Yayyy good job \c:/ If your teacher wants you to leave it as an exact value you would have:\[\large\rm x=\frac{\ln4.75}{2+\ln4}\] But yes, that's a correct decimal approximation.

  37. anonymous
    • one year ago
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    Yay!!! You are the BEST!!! Thanks so much!

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