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3^2x=5(3^x)+36 whats the exact value of x?

  1. Dec 7, 2004 #1
    I let A=3x and then i got (A-9)(A-4) when i factored the quadratic. Then when I got a common base of 3 I solved one x value to be x=2 but Im not sure about the other one? 3^x=4 what is the common base? I used the graphing calculator instead and got x=1.26 is this correct? could i have done it another way? :rolleyes:
     
  2. jcsd
  3. Dec 7, 2004 #2

    Zurtex

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    A=3^x not 3x right?

    I've not checked your algebra but assuming it is right:

    [tex]3^x = 4[/tex]

    [tex]x = \log_3 4[/tex]

    or:

    [tex]3^x = 4[/tex]

    [tex]\ln \left( 3^x \right) = \ln (4)[/tex]

    [tex]x \ln (3) = \ln(4)[/tex]

    [tex]x = \frac{\ln(4)}{\ln(3)}[/tex]

    Edit: You have made a mistake, check your pluses and minus when you factored again, there is only one real root.
     
    Last edited: Dec 7, 2004
  4. Dec 7, 2004 #3
    Thanks I got it :smile:
     
  5. Dec 7, 2004 #4

    dextercioby

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    Final answer,...............................................YES.But the equation subject of this post will not appear in the solution to your problem,as the decomposition u found was wrong.It should have stated (A-9)(A+4) with only the "9" sollution adittable,since the exponetial of real numbers cannot be negative.
    Anyway,for the equation u stated,here goes my say:
    U got [itex] 3^x =4 [/itex].Apply logarithm in any base on both sides of the eq.Apply it in the base 3 to find:[itex] x=\log_{3} 4 [/itex].Apply in the "natural" basis to find:[itex] x=\frac{\ln 4}{\ln 3} [/itex].
    Take the pocket intelligent calculator (the one with scientifical functions) and compute the ratio between the 2 natural logarithms and you"ll find "x" with a staggering approximation (actuatlly the calculator's number of possible decimals).

    Equations like the one u encountered (the last one) are called transcendental equations and,in general,solutions are transcendental numbers,i.e. numbers like "e" and "pi".

    Daniel.

    PS.Edit:Zurtex,i hadn't seen yor post,as i would have cut mine to half,sice you dealt with the transcedental eq.
     
    Last edited: Dec 7, 2004
  6. Dec 7, 2004 #5
    Since one value was negative I think there is no solution for that one since when a positive base is raised to any exponent the result should be a positive value. We are not doing log this year, thanks for ur help!
     
  7. Dec 7, 2004 #6

    dextercioby

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    I edited my post.I assumed that the decomposition u found was correct (and so did Zurtex),so i proceded with the unclarities regarding the last equation.
    Now my post makes perfect sense.Almost... :tongue2:
     
  8. Dec 7, 2004 #7
    I dont understand am I wrong?
     
  9. Dec 7, 2004 #8

    dextercioby

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    YES,but,thankfully u figured out the mistake.

    Daniel.
     
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