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anonymous

  • one year ago

Given the parent functions f(x) = 5x − 1 and g(x) = 3x − 9, what is g(x) − f(x)?

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  1. UsukiDoll
    • one year ago
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    given: g(x) = 3x-9 f(x) = 5x-1 and we need to solve g(x) - f(x)

  2. UsukiDoll
    • one year ago
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    so we plug these functions into g(x)-f(x) 3x-9-(5x-1) now distribute the negative on the right hand side of the equation

  3. anonymous
    • one year ago
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    3x-9(-5x+1)?

  4. UsukiDoll
    • one year ago
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    ok... that's good 3x-9-5x+1 now we combine like terms.. we can also rearrange this equation to make it easier to solve so instead of 3x-9-5x+1 we can rewrite it as 3x-5x-9+1

  5. anonymous
    • one year ago
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    oh the equation is 3^x

  6. anonymous
    • one year ago
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    im sorry

  7. UsukiDoll
    • one year ago
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    ahhhhhhhh! ok no problem... just rewrite what f(x) and g(x) was supposed to be

  8. UsukiDoll
    • one year ago
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    ok was g(x) = 3^x-9 ?

  9. anonymous
    • one year ago
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    f(x)=5x-1 g(x)=3^x-9

  10. anonymous
    • one year ago
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    @jim_thompson5910

  11. UsukiDoll
    • one year ago
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    ok we do the same process again g(x) -f(x) 3^x-9-(5x-1)

  12. UsukiDoll
    • one year ago
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    actually we have already done the distribution of the negative, 3^x-9-5x+1 now combine like terms for -9+1

  13. anonymous
    • one year ago
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    -8

  14. UsukiDoll
    • one year ago
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    \[3^x-9+1-5x\]

  15. UsukiDoll
    • one year ago
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    yup \[3^x-8-5x\]

  16. UsukiDoll
    • one year ago
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    since we can't combine any more terms, that's the answer

  17. anonymous
    • one year ago
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    Given the parent functions f(x) = log10 x and g(x) = 5x − 2, what is f(x) • g(x)? f(x) • g(x) = log10 x5x − 2 f(x) • g(x) = log10 (5x − 2)x f(x) • g(x) = 5x log10 x + 2 log10 x f(x) • g(x) = 2 log10 x − 5x log10 x

  18. anonymous
    • one year ago
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    let me type the equation correctly

  19. anonymous
    • one year ago
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    \[f(x)=\log_{10} x \] g(x)=5x-2

  20. anonymous
    • one year ago
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    firt answer is log10 x^5-2 second is log10(5x-2)^x

  21. anonymous
    • one year ago
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    @jim_thompson5910

  22. UsukiDoll
    • one year ago
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    we are using multiplication \[f(x) \cdot g(x) \] once again we place our f(x) function and g(x) into the formula

  23. UsukiDoll
    • one year ago
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    \[\log10_x[5x-2]\] use distribution

  24. anonymous
    • one year ago
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    what do you distribute

  25. UsukiDoll
    • one year ago
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    distribute the log

  26. UsukiDoll
    • one year ago
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    \[\log_{10}x[5x-2]\]

  27. anonymous
    • one year ago
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    i dont have a graphing or scientific calc to solve

  28. UsukiDoll
    • one year ago
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    we can just distribute the log |dw:1435728843362:dw|

  29. anonymous
    • one year ago
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    x(log10 5x-log10 2)?

  30. UsukiDoll
    • one year ago
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    |dw:1435728950642:dw| now distribute the log with the negative -2

  31. UsukiDoll
    • one year ago
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    distribution is multiplication what's that log times -2?

  32. anonymous
    • one year ago
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    idk i dont have a calc

  33. UsukiDoll
    • one year ago
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    we don't need a calculator this time

  34. UsukiDoll
    • one year ago
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    \[\log_{10}x[5x-2] \rightarrow \log_{10}x[5x]-\log_{10}x[2]\] based on your choices, it doesn't require you to calculate the log portion

  35. anonymous
    • one year ago
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    so its either c or d?

  36. anonymous
    • one year ago
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    wouldnt it be d because its - and not +?

  37. UsukiDoll
    • one year ago
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    I'm not sure.. @jim_thompson5910 can help out.. I need to go

  38. anonymous
    • one year ago
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    @jim_thompson5910

  39. jim_thompson5910
    • one year ago
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    This question right? Given the parent functions f(x) = log10 x and g(x) = 5x − 2, what is f(x) • g(x)? f(x) • g(x) = log10 x5x − 2 f(x) • g(x) = log10 (5x − 2)x f(x) • g(x) = 5x log10 x + 2 log10 x f(x) • g(x) = 2 log10 x − 5x log10 x

  40. anonymous
    • one year ago
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    yes and i corrected how it is right under neath that

  41. jim_thompson5910
    • one year ago
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    \[\Large f(x) = \log_{10}(x)\] \[\Large g(x) = 5x-2\] You will distribute the "log(x)" portion through like so \[\Large f(x) \cdot g(x) = \log_{10}(x)\cdot(5x-2)\] \[\Large f(x) \cdot g(x) = \log_{10}(x)\cdot 5x + \log_{10}(x)\cdot(-2)\] \[\Large f(x) \cdot g(x) = 5x\log_{10}(x)-2\log_{10}(x)\]

  42. anonymous
    • one year ago
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    so d?

  43. anonymous
    • one year ago
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    its backwards but

  44. jim_thompson5910
    • one year ago
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    can you post a screenshot of the full thing? some symbols didn't make it (at least I think?)

  45. anonymous
    • one year ago
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  46. jim_thompson5910
    • one year ago
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    D is close but not quite 100%

  47. jim_thompson5910
    • one year ago
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    there has to be a typo that they made

  48. anonymous
    • one year ago
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  49. anonymous
    • one year ago
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    could it be c because when you multiply its equivalent to adding with exponets or something like that lol?

  50. jim_thompson5910
    • one year ago
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    hints: * look at rule 2 at the link below http://www.purplemath.com/modules/logrules.htm * factor 5x-5 and you'll see something cancel

  51. anonymous
    • one year ago
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    i still dont understand please help its one in the morning

  52. jim_thompson5910
    • one year ago
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    \[\Large f(x) = \log_{3}(5x-5)\] \[\Large g(x) = \log_{3}(x-1)\] Use rule 2 from that link \[\Large f(x)-g(x) = \log_{3}(5x-5)-\log_{3}(x-1)\] \[\Large f(x)-g(x) = \log_{3}\left(\frac{5x-5}{x-1}\right)\] \[\Large f(x)-g(x) = \log_{3}\left(\frac{5(x-1)}{x-1}\right)\] \[\Large f(x)-g(x) = \log_{3}\left(\frac{5\cancel{(x-1)}}{\cancel{x-1}}\right)\] \[\Large f(x)-g(x) = \log_{3}\left(5\right)\]

  53. anonymous
    • one year ago
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    i thought we were on the other one still lol but yeah it make since now that i know what were talking about

  54. jim_thompson5910
    • one year ago
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    oh the other one has a typo I think

  55. anonymous
    • one year ago
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    which would you pick

  56. jim_thompson5910
    • one year ago
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    D I guess since it's close just off by a sign

  57. anonymous
    • one year ago
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    it was wrong i guess it was c

  58. anonymous
    • one year ago
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    thanks thoough!

  59. jim_thompson5910
    • one year ago
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    you'll have to tell your teacher about the typo

  60. UsukiDoll
    • one year ago
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    I second on that typo comment. Even I was in ??????????? land when I found out that there was a sign switch on C and D. It's like we have the answer, but the choices they gave messed up on the signs.

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