Solve
log(2x)+ 3 = 4
x=5
What are two points for the following function?
f(x)= 3x
(0, 1)
(1, 3)
Find vertex and a.o.s.
f(x)= -6x2 +12 -1
(1, 5)
x=1
Evaluate
log7(1)=
ln(e)=
log9(3)=
0
1
1/2
log2(3x-4) = 3
x=4
Describe translations. What is the asymptote?
f(x)= 3x-1 + 2
y=2
Move right one unit
Move up two units
x - 51/2 +6=0
x=9
x=4
Determine the number of solutions for the following
3x2 -4x +3=0
2 imaginary solutions
or NO real solution
Solve
2 - log7x = log7(x-48)
x= 49
Write in logarithmic form.
122=144
log12 144 = 2
SOLVE SYSTEM
-5x - 4y = 2
4x + y = 5
(2, -3)
Solve
5x-2=7
x=ln7/ln5 + 2
Solve
ln(x-4) = ln(x+6) - lnx
x= 6
x=-1 is no good
f(x)= log2(x)
(1, 0)
(2, 1)
x=0
f-1(x)= 2x
Given
f(x)= x2+2x and g(x)=x+2
Find (f o f)(3)
255
Write as single log
1/2lnx + ln(x2+1) - ln(x+1)
ln ( x1/2(x2+1) ) / (x+1)
Solve
log2(4x+10) = 3
x= -1/2
Describe translations of h(x).
What is the asymptote?
h(x)= log2(x+1) + 2
Move left one unit
Move up two units
x= -1
Given
f(x)= x2+2x and g(x)=x+2
Find (g o f)(3)
17
42x+3=8x
x= -6