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anonymous

  • one year ago

i need someone to walk me through solving this equation: 200x + 1,000 = 10(1.5^x) medal and fan.(:

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  1. anonymous
    • one year ago
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    Is that 1.5 to the x power?

  2. anonymous
    • one year ago
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    yes

  3. sweetburger
    • one year ago
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    lol i plugged this into my calculator and it took about a min to solve for an x value

  4. anonymous
    • one year ago
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    haha

  5. sweetburger
    • one year ago
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    well first i know we might aswell divide everything by 10 to get rid of it

  6. Nnesha
    • one year ago
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    first of all divide both sides by 10 so you can take ln both sides.......-....

  7. anonymous
    • one year ago
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    okay

  8. anonymous
    • one year ago
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    one second please

  9. anonymous
    • one year ago
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    okay i got it

  10. anonymous
    • one year ago
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    sorry it took so long, i had to deal with something else.

  11. sweetburger
    • one year ago
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    you can factor out a 20 on the right side of the equation

  12. anonymous
    • one year ago
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    i have 200x + 100 = 1(1.5^x)

  13. anonymous
    • one year ago
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    why am i putting a twenty there?

  14. sweetburger
    • one year ago
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    ok so u didnt factor out the 10 from everything

  15. anonymous
    • one year ago
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    yes i divided 10 and 1000 by 10

  16. sweetburger
    • one year ago
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    should be 20x + 100 = 1.5^x

  17. anonymous
    • one year ago
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    wait what

  18. sweetburger
    • one year ago
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    divide everything by 10...

  19. anonymous
    • one year ago
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    i thought you only divide the sides that don't have x

  20. anonymous
    • one year ago
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    so 1000 and 10

  21. sweetburger
    • one year ago
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    what... u have confused me

  22. sweetburger
    • one year ago
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    just divide each value in the equation by 10 so 200/10 1000/10 and 10/10

  23. sweetburger
    • one year ago
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    200x/10 +1000/10 = 10/10(1.5^x) now solve that @grunge101

  24. anonymous
    • one year ago
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    i thought you weren't supposed to divide the numbers with x

  25. sweetburger
    • one year ago
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    why did u think that? remember that 200 in front of x is just a coefficient you still have to reduce it

  26. anonymous
    • one year ago
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    |dw:1434065203963:dw|

  27. Nnesha
    • one year ago
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    or first you can find common factor 200x+1000

  28. sweetburger
    • one year ago
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    @nelsonjedi made a good drawing of what you should do

  29. freckles
    • one year ago
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    in any case I think you are going to a numerical or graphing approach to approximate the solutions to the equation

  30. anonymous
    • one year ago
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    thank you @nelsonjedi

  31. sweetburger
    • one year ago
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    im getting some strange values for x when i solve for x on my calc

  32. anonymous
    • one year ago
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    okay so i'm a little it confused. sorry guys i keep having problems on my end . so first i divide everything by 10

  33. Nnesha
    • one year ago
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    yes right

  34. anonymous
    • one year ago
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    okay, then what?

  35. anonymous
    • one year ago
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    i have 20x + 10 = 1(1.5^x)

  36. freckles
    • one year ago
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    what method are you suppose to use to approximate the solutions? Or are are you familiar with lambert function?

  37. anonymous
    • one year ago
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    um, i don't really know. i just have to solve it.

  38. anonymous
    • one year ago
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    algebraically

  39. freckles
    • one year ago
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    and aren't you suppose to have: 200x + 1,000 = 10(1.5^x) this is equivalent to 20x+100=(1.5)^x

  40. anonymous
    • one year ago
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    oh yes, i meant 10.

  41. anonymous
    • one year ago
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    and why isn't the 1 in front of (1.5^x) ?

  42. freckles
    • one year ago
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    \[200x+1000=10(1.5)^x \\ \text{ divide both sides by 10} \\ \frac{200x+1000}{10}=\frac{10(1.5)^x}{10} \\ \frac{200x}{10}+\frac{1000}{10}=\frac{10}{10}(1.5)^x \\ 20x+100=1(1.5)^x \text{ you can put a 1 but \it is \not needed } \\ \text{ this is the same as } 20x+100=(1.5)^x \\ \]

  43. anonymous
    • one year ago
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    okay i understand.

  44. freckles
    • one year ago
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    so if you want to solve this algebraically you will have to know the lambert function

  45. anonymous
    • one year ago
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    okay, could you walk me through it?

  46. freckles
    • one year ago
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    are you sure you have to solve algebraically ?

  47. anonymous
    • one year ago
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    honestly, i have no idea. this is something i learned a long time ago, and i struggled with it then. i don't remember what the process to solve it was called.

  48. anonymous
    • one year ago
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    but i need to just solve this problem so i can understand how to do it and move on to the further step of this sort of equation using systems of equations approximately.

  49. freckles
    • one year ago
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    The equation you want to solve currently is done in most algebra classes by finding approximations to the solutions using numerical approach or graphing approach. I would graph both expressions (the left and right of the equal sign expressions) and see where they intersect. The intersection of these curves (line) are the solutions.

  50. anonymous
    • one year ago
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    well i don't want to solve it using approximation and systems of equations and whatnot. right now, i need to solve it a step backwards.

  51. anonymous
    • one year ago
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    but you know what, i don't really need help anymore. it doesn't matter. i'll figure it out later. you'll still get a medal for trying so hard and dedicating so much of your time tonight to helping me. thank you. :)

  52. freckles
    • one year ago
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    \[20x+100=(1.5)^x \\ \text{ or } 20x+100=(\frac{3}{2})^x \\\] \[20(x+5)=(\frac{3}{2})^x \\ \text{ \let } t=x+5 \\ 20t=(\frac{3}{2})^{t-5} \\ 20t(\frac{3}{2})^{5}=(\frac{3}{2})^t \\ 20(\frac{3}{2})^5=\frac{1}{t}(\frac{3}{2})^t \\ 20(\frac{3}{2})^5=t^{-1}(\frac{2}{3})^{-t} \\ \frac{1}{20}(\frac{2}{3})^5=t(\frac{2}{3})^t\\ \frac{1}{20}(\frac{2}{3})^5=t(e^{\ln(\frac{2}{3}) \cdot t}) \\ \ln(\frac{2}{3}) \frac{1}{20} (\frac{2}{3})^5=\ln(\frac{2}{3})te^{\ln(\frac{2}{3} )\cdot t} \\ \text{ take } W \text{ of both sides } \\ W(\frac{1}{20} \ln(\frac{2}{3}) (\frac{2}{3})^5)=\ln(\frac{2}{3}) \cdot t \\ \text{ so we have } t=\frac{1}{\ln(\frac{2}{3})} W(\frac{1}{20}\ln(\frac{2}{3}) (\frac{2}{3})^5) \] but recall t=x+5 \[x=\frac{1}{\ln(\frac{2}{3})}W(\frac{1}{20}\ln(\frac{2}{3})(\frac{2}{3})^5)-5\]

  53. freckles
    • one year ago
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    I post this thinking this is not what you really want and you are looking for an approach that is more likely to occur in algebra class which for this type of equation is a most likely some numerical/graphing type approach. But the above is I would consider very algebraic approach.

  54. anonymous
    • one year ago
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    thank you! @freckles i just learned how to solve it approximately.

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