Various Problems for Precalculus Exam, Unit 2

[jacksonpeeble]

Homework Statement

9.e. Solve the equation for x: x²*e^x+x*e^x-e^x=0

19. If log_xa=0.578 and log_ay=1.161, what is the value of log_xy?

26. Use the fact that log_b4=x and log_b7=y to evaluate log_b(64/49).

Homework Equations

9.e. x²*e^x+x*e^x-e^x=0

19. log_xa=0.578 and log_ay=1.161

26. log_xa=0.578 and log_ay=1.161

The Attempt at a Solution

9.e. Factored equation so e^x(x²+x-1), then plugged into quadratic equation so \frac{-1+-\sqrt{1^{2}-4*1*-1}}{2*1} which equals \frac{-1+-\sqrt{5}}{2}. 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*\frac{-1+-\sqrt{5}}{2}.

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.

 

[jgens]

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).

 

[jacksonpeeble]

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 ;-)

 

[jgens]

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?

 

[Дьявол]

The first one is correct since e^x can never be zero. There are 2 solutions.

2nd one:

\frac{log_ay}{log_ax}=log_xy

\frac{1}{log_ax}=0.578

log_ax=\frac{1}{0.578}

Now divide log_ay with log_ax

3nd one:

3log_b4=3x[/itex]<br /> <br /> log_b64=3x[/itex]&lt;br /&gt; &lt;br /&gt; 2log_b7=2y[/itex]&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; log_b49=2y[/itex]&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; log_b(64/49)=log_b64-log_b49&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; Now just substitute.&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; Regards.