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Exponent problem

  1. Apr 2, 2012 #1
    Solve for x,
    (73/10)^(x-1)+5 - 3^(x+1)=0
     
  2. jcsd
  3. Apr 2, 2012 #2

    jedishrfu

    Staff: Mentor

    we cant solve this for you, please show your attempts at solving it first so we can comment on where you're stuck or got it worng.
     
  4. Apr 2, 2012 #3
    I literally have no idea, tried log, ln, e^x,

    my first work:
    7.3^(x-1)+5=3^(x+1)

    log_(3) (7.3^(x-1)+5)=x+1

    e^(log_3 (7.3^(x-1)+5) / e^x = e

    (7.3^(x-1)+5)/e^x =e
    7.3^(x-1)+5= ee^x
    7.3^(x-1) -ee^x=-5

    my second work,
    7.3^(x-1)=3^(x+1)-5

    (x-1)log(73/10)=log(3^(x+1)-5)

    xlog(73/10)-log(73/10) = log(3^(x+1)-5)

    (e^(xlog(73/10)) / (e^(log73/10) = 3^(x+1)-5

    e^(xlog73/10) = 3^(x+1) e^(log73/10)-5e^(log73/10)
     
  5. Apr 2, 2012 #4

    Mentallic

    User Avatar
    Homework Helper

    Just to clarify, you mean
    [tex]\left(\frac{73}{10}\right)^{x-1}-3^{x+1}+5=0[/tex]

    right? Because I don't believe this expression can be solved algebraically. You can, of course, find a numerical solution to the problem however.
     
  6. Apr 2, 2012 #5

    jedishrfu

    Staff: Mentor

    I agree with Mentallic, if its as he wrote there is no simple algebraic solution only a numeric one.
     
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