Expert: Steve Holleran - 11/28/2007

Question
Okay- here's what I'm having touble with...
1. what's the equation of the line with point (3,-2) and perpendicular to the line 2x-5y=10

2. what's the vertex, focus, and directrix of the parabola with the equation y=1/8x squared

3. what's the solution to the equation 4 to the 5-x power=7 to the x power

4. solve 1n(2-x)+1n(5-x)=1n(37-x)

Thanks so much!

Answer
Hi Alisa,

1.  The given line has slope = -2/-5 = 2/5, so any perpendicular has slope -5/2.  If it passes through (3, -2), then we can use point-slope form:

   (y - -2) = -5/2 * (x - 3)

    y + 2  = -5/2 * (x - 3)  which becomes 5x + 2y = 11

2.  The form you want to use is

             y-k = (1/4p)(x-h)^2   where (h,k) is the vertex and p is the distance from the vertex to the focus and to the directrix.

Here, h and k are 0 , so the vertex is (0,0).

4p = 8, so p = 2, so the focus is at (0,2) and the directrix is y = -2.

3.  4^(5-x) = 7^x

Take the ln of both sides:

  ln[4^(5-x)] = ln[7^x]  

so  (5-x) * ln 4 = x * ln 7

   5* ln 4 - x* ln 4 = x * ln 7

   5 ln 4 = x ln 7 + x ln 4 = x(ln 7 + ln 4) = x ln 28

so x = 5 ln 4 / ln 28 = 2.080

4.  ln(2-x) + ln(5-x) = ln(37-x)

   ln[(2-x)(5-x)] = ln 37

so     (2-x)(5-x) = 37

     10 - 2x - 5x + x^2 = 37

     x^2 - 7x + 10 = 37  so  x^2 - 7x - 27 = 0

and x = 9.765 or -2.765

Steve