兩條數學問題,線方程加log方程,謝謝

the slope of equation is -3
the line of equation cut the mid point of (2,6) and (8,6)
mid point=((2+8)/2,(6+6)/2]
mid point=(5,6)
[(y-6)/(x-5)]=-3
y-6=-3x+15
y+3x-21=0
the line of equation is y+3x-21=0.

log(3x)^4 - logx^2 = log81 - logx
log(3x)^4 - logx^2+logx=log81
log([(3x)^4](x)/x^2)=log81
log(81x^3)=log81
log81+3logx=log81
3logx=0
logx=0
x=1

2013-01-03 19:27:48 補充：
SOR 我黎過 英文唔好....
the slope of equation 乘 -3=-1
the slope of equation=1/3
the line of equation cut the mid point of (2,6) and (8,6)
mid point=((2+8)/2,(6+6)/2]
mid point=(5,6)
[(y-6)/(x-5)]=1/3
3y-18=x-5
3y-x-13=0
the line of equation is 3y-x-13=0

log3x^4 - logx^2 = log81 - logx

2013-01-03 19:32:02 補充：
log3x^4 - logx^2+logx= log81 [ logAB=logA+logB][logA/B=logA-logB]
log(3x^4/x^2)+logx= log81
log(3x^2)+ logx= log81
log[(3x^2)(x)]= log81
log[(3x^3)= log81
3x^3=81
x^3=27
x^3=3^3
x=3

slope perpendicular to -3
係 to x=-3, 定係 to y=-3, 定係 to 另一條 straight line with slope = -3 ?