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Homework Help: Various Problems for Precalculus Exam, Unit 2

  1. Jan 11, 2009 #1

    jacksonpeeble

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    1. The problem statement, all variables and given/known data
    9.e. Solve the equation for x: x2*ex+x*ex-ex=0

    19. If logxa=0.578 and logay=1.161, what is the value of logxy?

    26. Use the fact that logb4=x and logb7=y to evaluate logb(64/49).


    2. Relevant equations
    9.e. x2*ex+x*ex-ex=0

    19. logxa=0.578 and logay=1.161

    26. logxa=0.578 and logay=1.161


    3. The attempt at a solution
    9.e. Factored equation so ex(x2+x-1), then plugged into quadratic equation so [tex]\frac{-1+-\sqrt{1^{2}-4*1*-1}}{2*1}[/tex] which equals [tex]\frac{-1+-\sqrt{5}}{2}[/tex]. However, I'm wondering exactly what to do with the initial ex that I factored out - the answer key we were provided with does not include it at all, but I assumed that we would pin it in front of the equation so that our final answer is ex*[tex]\frac{-1+-\sqrt{5}}{2}[/tex].

    19. I remember doing these and that the solution is fairly obvious, I just forget what it is.

    26. Sorry, this is one where I have no idea.
     
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  3. Jan 11, 2009 #2

    jgens

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    For the first problem, think zero product property. You've already found two values that will yield a zero. Now, when does e^x equal 0?

    The solution to the second problem can be obtained by noting a = x^0.578 and y = a^1.161

    Assuming the desired answer need be expressed in terms of x and y, use the properties alog(x) = log(x^a), log(a/b) = log(a) - log(b).
     
  4. Jan 11, 2009 #3

    jacksonpeeble

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    Thanks for the reply!

    Ok, so I was fairly accurate with #9.

    For #19, I understand that - from the revised equations, how are we supposed to connect them into a single solution?

    #26 is still up for grabs ;-)
     
  5. Jan 11, 2009 #4

    jgens

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    Yeah, since e^x is never zero, you already had the two solutions.

    For 19 try a simple substitution.

    Alright, 64 = 4^3 and 49 = 7^2. Therefore, log(64) = 3x and log(49) = 2y. Can you take it from there?
     
  6. Jan 11, 2009 #5
    The first one is correct since ex can never be zero. There are 2 solutions.

    2nd one:

    [tex]\frac{log_ay}{log_ax}=log_xy[/tex]

    [tex]\frac{1}{log_ax}=0.578[/tex]

    [tex]log_ax=\frac{1}{0.578}[/tex]

    Now divide [itex]log_ay[/itex] with [itex]log_ax[/itex]

    3nd one:

    [tex]3log_b4=3x[/itex]

    [tex]log_b64=3x[/itex]

    [tex]2log_b7=2y[/itex]

    [tex]log_b49=2y[/itex]

    [tex]log_b(64/49)=log_b64-log_b49[/tex]

    Now just substitute.

    Regards.
     
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