Study Flashcards
Solve 25^x = 125
- (5^2)^x = 5^3
- 5^2x = 5^3
- ln(5^2x) = ln(5^3)
- 2x = 3
- x = 3/2
Solve 9^x = 3^x+1
- (3^2)^x = 3^x+1
- ln(3^2x) = ln(3^x+1)
- 2x = x+1
- x = 1
Solve 5^2x-3 = 3^x+1
- ln(5^2x-3) = ln(3^x+1)
- (2x-3)ln(5) = (x+1)ln(3)
- 2xln(5)-3ln(5) = xln(3)+ln(3)
- 2xln(5)-xln(3) = 3ln(5) + ln(3)
- x(2ln(5)-ln(3)) = 3ln(5)+ln(3)
- x = 3ln(5)+ln(3) / 2ln(5)-ln(3)
Solve 3^x - 8•3^-x = 2
- (3^x)^2 - 8 = 2•3^x 3^- x cancels
- Let u = 3^x
- (u^2) - 8 = 2u
- u^2 -2u-8 = 0
- (u-4)(u+2)
- 3^x = 4 or x = log3(4)
Solve log2(x-3) + log2(x-4) = 1
- log2[(x-3)(x-4)] = 1
- (x-3)(x-4) = 2
- x^2-7x+12=2
- (x-5)(x-2)
- x = 5
Solve log2(x+4) - log2(x+3) = 1
- log2[x+4/x+3] = 1
- x+4/x+3 = 2
- x+4 = 2x+6
- x = -2
How to get rid of this log?
log2(4x) = 3
With the base (2) to the x.
2^log2(4x) = 2^3
4x = 8
Solve 5(.7)^x + 3 < 18
- 5(.7)^x < 15
- .7^x < 3
- ln(.7^x) < ln(3)
- xln(.7) < ln(3)
- x > ln(3) / ln(.7)
Solve log(2x-5) <= 1
- 2x-5 > 0 so x > 5/2
- 2x-5 <= 10 so x <= 15/2
- (5/2, 5/12)
Write in long form:
ln[(x+5) / (x-1)√x+4]
4ln(x+5) - ln(x-1) - 1/2ln(x+4)
What does log3(1/27) equal?
-3
What does log64(4) equal?
1/3
If an inequality is decreasing on both sides, then
The inequality is flipped
Condensed form of:
9/2log(x)-log(3y)+log(2z)
log √x^9•2z / 3y
log2(x) = -3 equals
1/8
log3(x) = -1 equals
1/3
10^3x+5 = 11
ln(11)-5ln(10) / 3ln(10)
Two rays form an
Angle
Forms an angle
Two rays
Angle starts at and ends at
Initial side and terminal side
Initial side and terminal side
Sides that an angle starts and ends from
A counterclockwise angle is
Positive
Which angle is positive?
Counterclockwise
A clockwise angle is
Negative
Negative angle
Counterclockwise
Two angles with same initial and terminal sides
Coterminal