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For values of x which a solution exists solve for y

  1. Apr 8, 2010 #1
    1. The problem statement, all variables and given/known data
    (For those values of x for which a solution exists), solve the following equation for y
    3e^(2y−8) = (2x^5)−3

    2. Relevant equations
    1. xln e= x

    3. The attempt at a solution
    1.3e^(2y−8) = (2x^5)−3
    2.e^(2y−8) = ((2x^5)−3)/3
    3.(2y-8)ln e = (ln (2x^5)-3)/3
    4.(2y-8) = (ln (2x^5)-3)/3
    5.y= ((ln (2x^5)-3/6)+8

    this answer was wrong just wondering where i went wrong??
     
  2. jcsd
  3. Apr 8, 2010 #2
    [tex]e^{2y-8}=\frac{2x^{5}}{3}-1[/tex]

    [tex]2y-8=ln({\frac{2x^{5}}{3}-1})[/tex]

    [tex]2y=ln(\frac{2x^{5}}{3}-1)+8[/tex]

    [tex]y=\frac{ln(\frac{2x^{5}}{3}-1)}{2}+4[/tex]
     
  4. Apr 8, 2010 #3
    Thanks man
     
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