(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

9.e. Solve the equation for x: x^{2}*e^{x}+x*e^{x}-e^{x}=0

19. If log_{x}a=0.578 and log_{a}y=1.161, what is the value of log_{x}y?

26. Use the fact that log_{b}4=x and log_{b}7=y to evaluate log_{b}(64/49).

2. Relevant equations

9.e. x^{2}*e^{x}+x*e^{x}-e^{x}=0

19. log_{x}a=0.578 and log_{a}y=1.161

26. log_{x}a=0.578 and log_{a}y=1.161

3. The attempt at a solution

9.e. Factored equation so e^{x}(x^{2}+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 e^{x}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 e^{x}*[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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# Homework Help: Various Problems for Precalculus Exam, Unit 2

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