2^2-2x > 3
(-∞,-1) U (3, ∞)
(3x+3)/(2x-4)>0
(-∞,-1)U(2,∞)
sqrt(20-8x)=x
x=2
Graph and determine its doman, range and asymptote
f(x)= (1/3)^x
D: (∞ ,-∞)
R: (0, ∞)
Asymptote: y=0
Decreasing if
0<b<1
Expand the following expression as much as possible
log₂((sqrtx)/(36y^4))
1/3log₂x-2-4 log₂ y
2x^2- x > 5
(-∞,-3)U( 5/2 ,∞)
(x-1)/(x+1)≥0
(-∞,-1)U[1,∞)
sqrt(x+10)=x-2
x=6
Graph and determine its doman, range and asymptote
f(x)= 3^(x-4)
D: (∞ ,-∞)
R: (0, ∞)
Asymptote: y=0
-Right 4 translation
Increasing if
b >1
Write as a single logarithm
1/2(log _5x+log_5y)-2log_5(x+1)
log_5((sqrtxy)/(x+1)^2)
x^3+3x^2≤x+3
(-∞,-3]U[-1,1]
(4-2x)/(3x+4)≤0
(-∞, -4/3) U [2, ∞)
5^(3/2)-25=0
.^3sqrt(25)
Graph and determine its doman, range and asymptote
f(x)= -3^(x+1)+2
D: (∞ ,-∞)
R: (2, -∞)
Asymptote: y=2
-Relfection across x-axis
-2 up translation
-1 left translation
Solve
log_2x+log_2(x-7)=3
x=8
x^3+x^2≤4x+4
(-∞,-2]U[-1,2]
(x+1)/(x+3)<2
(-∞,-5) U (-3, ∞)
x^(2/3)-x^(1/3)-6
x= 27 and x=-8
Slove the following exponential equations
1) 27^(x+3)=9^(x-1)
2)5^(3x-6)=125
1)x=-11
2)x=3
Solve
ln(x+2)-ln(4x+3)=ln(1/x)
Hint: apply the one-to-one property of logarithms
x=3
x^3≥9x^2
{0}U[9,∞)
(x+4)/(2x-1)≤3
(-∞,1/2)U[7/5, ∞)
7x^(2/3)-55x^(1/3)-8=0
x=512 and x=-1/343
Solve
e^(2x)-4e^x+3=0
x=ln3, 0
Solve
log_2(x-6)+log_2(x-4)-log_2x=2
x=12