13. Solve completely v x E R, show the equation in set and graphical notation:
3x-5/x-2 less thn or equal to 5/x+5
14. Solve v x E R: 2^x=7
15. If f(x)=2x-3, find f(log 10).
16. Simplify v x E R and give parameters: x^2+x-6/x-2
For #13, we would get everything on one side of the inequality. So, for example, subtracting 5/(x+5) from each side:
(3x-5)/(x-2) - 5/(x+5) <-- 0
Now, from here, this inequality doesn't exist at 2 points, when x = 2 and -5. So, we use those as "markers" to check other numbers. Also, we would write this as one fraction on the left, because the top would become a quadratic. We need to find the zeros for that also. The top would become:
(3x-5)(x+5) - 5(x-2)
3x^2 + 15x - 5x - 25 - 5x + 10
3x^2 + 5x - 15
The zeros for this are x = -3.22 and 1.55
So, we have 4 numbers that can break up or domain, 2, -5, -3.22, and 1.55. We use these to break up our domain into 5 sections:
All numbers less than -5
All numbers from -5 to -3.22
All numbers from -3.22 to 1.55
All numbers from 1.55 to 2
And, all numbers greater than 2
We pick a number from each section to determine if that section provides solutions, is a solution to the inequality. For instance, we can pick x = -6. Doing that:
(3(-6)-5)/(-6-2) - 5/(-6+5) <-- 0
-23/-8 - (-5) <-- 0
23/8 + 5 <-- 0
This is obviously false. So, this set of real numbers has no solution for the inequality. Doing the same for the other sections, we get where the solutions are:
from -5 to -3.22
and 1.55 to 2
Or, using parenthesis and brackets, we would have
(-5,-3.22] U [1.55,2)
Parenthesis because the inequality doesn't exist at -5 and 2. Brackets because the inequality does exist at -3.22 and 1.55, specifically being where the inequality = 0.
For 14, we can rewrite the equation as a log:
log(base 2) of 7 = x
Using the change of base formula, we can write:
(log 7)/(log 2) = x which is approx. 2.81. Assuming the base of the log is 10.
For 15, assuming the base of the log is 10, log 10 = 1. So, we are looking for f(1). f(1) = 2*1 - 3 = -1
For 16, we would factor the top into (x+3)(x-2). So, the "x-2" would cancel out, leaving x+3. However, the condition being x can't be 2. For, if x = 2, then we have 2-2 on the bottom = 0. And, we can't divide by 0.
I hope these help, Steve. Good luck.