# Homework Help: Various Problems for Precalculus Exam, Unit 2

1. Jan 11, 2009

### jacksonpeeble

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 $$\frac{-1+-\sqrt{1^{2}-4*1*-1}}{2*1}$$ which equals $$\frac{-1+-\sqrt{5}}{2}$$. However, I'm wondering exactly what to do with the initial 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*$$\frac{-1+-\sqrt{5}}{2}$$.

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.

2. Jan 11, 2009

### 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).

3. Jan 11, 2009

### 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 ;-)

4. Jan 11, 2009

### 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?

5. Jan 11, 2009

### Дьявол

The first one is correct since ex 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:

$$3log_b4=3x[/itex] [tex]log_b64=3x[/itex] [tex]2log_b7=2y[/itex] [tex]log_b49=2y[/itex] [tex]log_b(64/49)=log_b64-log_b49$$

Now just substitute.

Regards.