1)
If the simultaneous equatons: y=mx-1 and xy=1 have no real solutions,
find the range of values of m.
ans: m>-1/4
2)solve log(x+1)-log(3x^2-5)=-1
ans: 4.46
3)solve log(10^2x-56)-x=0
ans: 0.90
pls show clear steps.
thanks a lot.
F.5 more about equations
2010-09-11 2:57 am
回答 (2)
2010-09-11 3:24 am
✔ 最佳答案
As follows:圖片參考：http://i707.photobucket.com/albums/ww74/stevieg90/13-23.jpg
圖片參考：http://i707.photobucket.com/albums/ww74/stevieg90/14-22.jpg
2010-09-11 3:40 am
(1)
y=mx-1 ----(i)
xy=1 ----(ii)
put (i) into (ii)
x(mx-1)=1
mx^2-x-1=0
for there is no real solution, Δ<0
Δ=(-1)^2-4(m)(-1)<0
1+4m<0
m<-1/4 [it seems that the ans given got some mistake]
(2)
log(x+1)-log(3x^2-5)=-1
log[(x+1)/(3x^2-5)]=-1
(x+1)/(3x^2-5)=0.1
x+1=0.3x^2-0.5
0.3x^2-x-1.5=0
3x^2-10x-15=0
x=[10±√100-4(3)(-15)]/6
=(5+√70)/3=4.46 for x >-1
(3)
log(10^2x-56)-x=0
10^2x -56 = 10^x
10^2x - 10^x - 56 =0
10^x = (1±15) / 2
10^x = 8 ( x> (log56)/2 )
x=log8 = 0.9031
y=mx-1 ----(i)
xy=1 ----(ii)
put (i) into (ii)
x(mx-1)=1
mx^2-x-1=0
for there is no real solution, Δ<0
Δ=(-1)^2-4(m)(-1)<0
1+4m<0
m<-1/4 [it seems that the ans given got some mistake]
(2)
log(x+1)-log(3x^2-5)=-1
log[(x+1)/(3x^2-5)]=-1
(x+1)/(3x^2-5)=0.1
x+1=0.3x^2-0.5
0.3x^2-x-1.5=0
3x^2-10x-15=0
x=[10±√100-4(3)(-15)]/6
=(5+√70)/3=4.46 for x >-1
(3)
log(10^2x-56)-x=0
10^2x -56 = 10^x
10^2x - 10^x - 56 =0
10^x = (1±15) / 2
10^x = 8 ( x> (log56)/2 )
x=log8 = 0.9031
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