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
Coterminal
Two angles with same initial and terminal sides
An angle lies in a quadrant if its terminal side is
In that quadrant
If a terminal side is in a quadrant then so is
The angle
An acute angle
Has measure between 0 and 90
Has measure between 0 and 90
Acute angle
An obtuse angle has measure
90 and 180
Measure 90 and 180
Obtuse angle
A straight angle has measure
180
Has measure 180
Straight angle
Angle who’s vertex is center of a circle
Central angle
Central angle
Angle who’s vertex is center of a circle
Circumference of unit circle
2π or 360
2π or 360
Circumference of unit circle
Degrees to radians
π/180
π/180
Degrees to radians
Radians to degrees
180/π
180/π
Radians to degrees
Find the angle θ with 0 < θ < 2π
which is coterminal with
19π/4
- 19π/4 - 2(2π)
- 19π/4 - 4π
- 3π/4
The length S of an arc subtended by an angle of θ radius in a circle of radius r is
S = rθ
S = rθ
The length S of an arc subtended by an angle of θ radius in a circle of radius r is
Linear velocity is
Distance over time:
v = S/t
v = S/t
Linear velocity
Angular velocity is
Displacement over time:
ω = θ/t
ω = θ/t
Angular velocity
Revolutions / min is
Multiplied by 2π
csc θ =
hypotenuse / opposite
hypotenuse / opposite
csc θ
sec θ =
hypotenuse / adjacent
hypotenuse / adjacent
sec θ
cot θ
adjacent / opposite
adjacent / opposite
cot θ
Two triangles that contain θ are
Similar
Similar triangles
Contain angle measure of θ
tan θ relationship to sin θ and cos θ
tan θ = sin θ/cos θ
If sin θ = a/c then sin^2 θ =
a^2/b^2
Pythagorean identitity
sin^2 θ + cos^2 θ = 1
sin^2 θ + cos^2 θ = 1
Pythagorean identitity
Solve:
(2/5)^2 + cos^2 θ = 1
- 4/25 + cos^2 θ= 1
- cos^2 θ = 21/25
- cos θ = √21/5
Cofunction identities
sin θ = cos(π/2 - θ)
tan θ = cot(π/2 - θ)
sec θ = csc(π/2 - θ)
and other way around
sin θ = cos(π/2 - θ)
tan θ = cot(π/2 - θ)
sec θ = csc(π/2 - θ)
and other way around
Cofunction identities
