think i know how to do it i think you use distance formula and using coordinates of (c,log2(c), (c,log2(c-3) and using distance being equal to 4 and solve
think i know how to do it i think you use distance formula and using coordinates of (c,log2(c), (c,log2(c-3) and using distance being equal to 4 and solve
Ok, so since P has equation of y=log2(x) and the x coordinate of P is C , we sub c into y=log2(c) to get coordinate of P which is (C,LOG2(C), NOW FOR Q, since Q has x coordinate of c, and has equation of y=log2(x-3), we sub c in to get y and we get (c, log2(c-3)) they said the distance is 4 and we find distance between those two points using the distance formula and then solve for c
Ok, so since P has equation of y=log2(x) and the x coordinate of P is C , we sub c into y=log2(c) to get coordinate of P which is (C,LOG2(C), NOW FOR Q, since Q has x coordinate of c, and has equation of y=log2(x-3), we sub c in to get y and we get (c, log2(c-3)) they said the distance is 4 and we find distance between those two points using the distance formula and then solve for c
A rough sketch would help. Both points have x=c as the x coordinate, so the distance is simply the vertical distance, so the difference of two logs ...?
Ok, so since P has equation of y=log2(x) and the x coordinate of P is C , we sub c into y=log2(c) to get coordinate of P which is (C,LOG2(C), NOW FOR Q, since Q has x coordinate of c, and has equation of y=log2(x-3), we sub c in to get y and we get (c, log2(c-3)) they said the distance is 4 and we find distance between those two points using the distance formula and then solve for c
A rough sketch would help. Both points have x=c as the x coordinate, so the distance is simply the vertical distance, so the difference of two logs ...?
yeah i mean i didnt need to sketch the diagram, it was already given to us i was just overcomplicating the question, thanks for helping though