Powers & Roots Flashcards
2^3
2^3 = 8
2^4
2^4 = 16
2^5
2^5 = 32
2^6
2^6 = 64 ( = 4^3 )
2^7
2^7 = 128
2^8
2^8 = 256 ( = 4^4 )
2^9
2^9 = 512
3^3
3^3 = 27
3^4
3^4 = 81
4^3
4^3 = 64 ( = 2^6 )
4^4
4^4 = 256 ( = 2^8 )
5^3
5^3 = 125
5^4
5^4 = 625
6^3
6^3 = 216
7^3
7^3 = 343
8^3
8^3 = 512
9^3
9^3 = 729
(-1/2)^3
-1/8
(-1/2)^6
1/64
Is x^7 > x^6 ?
no clear answer: would be true for positive nb greater than one, false for negatives
also, if x=0, x^7 = x^6 = 0
If x < 1 , and x is unequal to 0, is x^7 > x^6?
If 0 < x < 1, and x is unequal to 0, is x^7 > x^6?
If x is any negative nb, then x^7 is negative and x^6 is positive, and any positive is greater than any negative; therefore, x^7 < x^6
NO
let's look at powers of x = 2/3 (2/3)² = 4/9 (2/3)^3 = 8/27 (2/3)^4 = 16/81 x² > x^3 > x^4 > x^6 > x^7 NO
(a^n)*(a^m)
(a^n)*(a^m) = a^(n+m)
a^m / a^n
a^m / a^n = a^(m-n)
a^0
a^0 = 1 if a is unequal to 0
a^3 / a^3 = a^0 = 1
(a^m)^n
(a^m)^n = a^(m*n)
b^(-n)
b^(-n) = 1/(b^n)
24x^(12)y^(9) / 18x^(-4)y^(3)
24x^(12)y^(9) / 18x^(-4)y^(3)
4x^(12)y^(9) / 3x^(-4)y^(3)
4x^(12)y^(6) / 3x^(-4)
(4/3)x^(16)y^(6)
Rank the following from smallest to biggest
I. (1/3)^(-8)
II. 3^(-3)
III. (1/3)^5
III, II, I
I. (1/3)^(-8) = 3^8 (approximately = 80² = 6400)
II. 3^(-3) = (1/3)^3 = 1/27
III. (1/3)^5 < (1/3)^3
(ab)^n
(a^n)(b^n)
(a/b)^n
a^n / b^n
(x²y^3)^4 / x^(5)y^(-5)
x^(8)y^(12) / x^(5)y^(-5)
x^(3)y^(17)
Factor out
17^30 + 17^20
(17^20)(17^10) + (17^10)(1)
17^20)(17^10 + 1
Simplify
3^32 - 3^28
(3^28)(3^4) - (3^28)(1)
(3^28)(3^4 - 1)
(3^28)(81 - 1)
80(3^28)
if b^s = b^t, then …
s = t
how to tackle units digit question
only consider single-digit products
57^123
7^1 = 7 (*7) 7^2 = ...9 (*7) 7^3 = ...3 (*7) 7^4 = ...1 (*7) 7^5 = ...7 7^6 = ...9 7^7 = ...3 7^8 = ...1 period of the pattern : 4 (repeats every 4) every time we get to a power of 4, we go back to the same place, unit digit of 1 so 7^120 = ...1 7^121 = ...7 7^122 = ...9 7^123 = ...3 and so : 57^123 = ...3
Is √2 > 1?
A) YES
B) NO
C) can’t be determined
A
the √ is printed in the question, so we consider only the positive square roots of 2
Let k² = 2. Is k > 1 ?
A) YES
B) NO
C) can’t be determined
C
we have to consider both the positive and negative square roots of 2
if q >or = 0, then √q …
√q > or = 0
√0 =?
0
if √A = B then …
A > or = 0 and B > or = 0
if A < B < C, then …
√A < √B < √C
√2
1.4
√3
1.7
√5
2.2
If (x - 3)² = 16, solve for x
x - 3 = 4 OR x - 3 = -4
x = 7 OR x = -1
If 1 < b, then b ?? b² and √b ?? b
If 1 < b, then b < b² and √b < b
If 0 < b < 1, then b ?? b² and √b ?? b
If 0 < b < 1, then b > b² and √b > b
If 7K is a positive integer, and if √K > K, then is K an integer?
A) YES
B) NO
C) can’t be determined
√K > K only happens when 0 < K < 1
can’t be an integer, would be either 1/7 or some multiple
B
2^3
8
3^3
27
4^3
64 = 2^6 = (2^3)²
5^3
125
6^3
216
7^3
343
8^3
512 = 2^9
9^3
729
10^3
1000
If 0 < b < 1 and if n > m, then …
0 < b < (m)√b < (n)√b < 1
the higher the order of the root, the closer the result is to 1
√PQ =?
√(P/Q) =?
√PQ = (√P)(√Q) √(P/Q) = √P / √Q
Simplify (√12)(√27)
√(1227) = √(1239) = √(369) = 6*3 = 18
Simplify √(4/49) and √(4/50)
√(4/49) = √4 / √49 = 2/7 √(4/50) = √(2/25) = √2/5
√64
8
√81
9
√121
11
√144
12
√169
13
√196
14
√225
15
Simplify √75, √12, √63, √80, √2800
√75 = √25*3 = 5√3 √12 = √3*4 = 2√3 √63 = √7*9 = 3√7 √80 = √16*5 = 4√5 √2800 = √28*100 = 10√28 = 10√4*7 = 10(2√7) = 20√7
If N = (2^6)(3^5)(5²)(7), express √N in simplified form
√N = (√2^6)(√3^5)(√5²)(7) √N = (√2^3*2^3)(√3²3²3^1)(√5²)(√7) √N = (2^3)(3²√3)(5)(√7) √N = 40(9√3)(√7) √N = 360√21
√72 - √32 =?
√72 - √32 = √236 - √216 = 6√2 - 4√2 = 2√2
(3√5)(7√2) =?
(3√5)(7√2) = 21(√5*√2) = 21√10
(3√5)(2√15) =?
(3√5)(2√15) = 6√515 = 6√553 = 65√3 = 30√3
(2√42)*(4√63) =?
(2√42)(4√63) = 8√4263 = 8√(6779) = 8√(237733) = 837√23 = 168√6
(54√35) / (18√5) =?
(54√35) / (18√5) = (54/18)*(√35/√5) = 3√7
(5√6)² =?
5²(√6)² = 25*6 = 150
(2√3)^4 =?
(2√3)^4 = 2^4 * (√3)^4 = 1633 = 169 = 144
OR
(2√3)^4 = ((2√3)²)² =(43)² = 12² = 144
(√2)^48 =?
((√2)²)^24 = 2^24
Is it always true that √(k²) = k?
NO
k = -4
k² = 16
√16 = 4
Solve for x
√(x + 3) = x - 3
√( x + 3 ) = x - 3 x + 3 = (x - 3)² x+3 = x² - 6x + 9 x² - 7x + 6 = 0 (x - 6)(x - 1) x = 1 or x = 6
Check x = 1
√(1+3) = 2
x - 3 = 1 - 3 = -2
doesn’t work
Check x = 6
√(6+3) = 3
x - 3 = 6 - 3 = 3
works
x = 6 is the only solution
Solve for x
√(2x - 2) = √(x - 4)
√(2x - 2) = √(x - 4)
2x - 2 = x - 4
x = -2
when we plug this in, this results in the square root of a negative on both sides
This equation has no solution
solve for x
2 + √(4 - 3x) = x
2 + √(4 - 3x) = x √(4 - 3x) = x - 2 4 - 3x = (x - 2)² 4 - 3x = x² - 4x + 4 x² - x = 0 x(x - ) = 0 x = 0 or x = 1
Check x =0
2 + √(4 - 3x) = 2 + √4 = 4
doesn’t work
Check x = 1
2 + √(4 - 3x) = 2 + √(4 - 3) = 3
doesn’t work
NO SOLUTION
2^(1/2) = K
(2^(1/2))² = K² K² = 2 K = √2 2^(1/2) = √2
2^(1/3) = K
(2^(1/3))^3 = K^3 K^3 = 2 K = (3)√2 2^(1/3) = (3)√2
b^(1/m) =?
(m)√b
2^(3/5) =?
2^(3/5) = (2^3)^(1/5) = (5)√(2^3) = (5)√8
OR
2^(3/5) = (2^(1/5))^3 = ((5)√2)^3
8^(4/3) =?
8^(4/3) = (3)√(8^4) = ((3)√8)^4 = 2^4 = 16
for positive nb
b^(1/2) =?
b^(m/n) =?
b^(1/2) = √b b^(m/n) = (b^m)^(1/n) = (b^(1/n))^m
If 7^(2x) = 7^(6 - x), then solve for x
2x = 6 - x 3x = 6 x = 2
If 49^x = 7^(6 - x), then solve for x
49^x = 7^(6 - x) (7²)^x = 7^(6 - x) 7^2x = 7^(6 - x) 2x = 6 - x 3x = 6 x = 2
to solve exponential equations, we must get equal bases on both sides. This may involve expressing the given bases as powers of similar bases
once the bases on both sides are equal, we can equate the exponents and solve
If ((5)√3)^(3x + 7) = 3^(2x), find x
(3^(1/5))^(3x + 7) = 3^(2x) 3^((3x + 7)/5) = 3^(2x) (3x + 7)/5 = 2x 3x + 7 = 10x 7 = 7x x = 1
to solve exponential equations, we must get equal bases on both sides. This may involve expressing the given bases as powers of similar bases
once the bases on both sides are equal, we can equate the exponents and solve
If 27^(2x - 2) = 881^(x + 1), find the value of x
27^(2x - 2) = 1^(x + 1) (3^3)^(2x - 2) = (3^4)^(x + 1) 3^(3(2x - 2)) = 3^(4(x + 1)) 6x - 6 = 4x + 4 2x = 10
to solve exponential equations, we must get equal bases on both sides. This may involve expressing the given bases as powers of similar bases
once the bases on both sides are equal, we can equate the exponents and solve
x = 5
Rationalize the following : 1/(√5) 2/(√3) 14/(√21) ( 4 - √6 ) / 2√3 2 / (√5 - 1) 8 / (√7 - √3) (4 + 2√5) / (3 + √5)
If the fraction has a single root in the denominator, we rationalize by multiplying y that root over itself
1/(√5) = (√5)/5
2/(√3) = (2√3)/3
14/(√21) = (14√21)/21 = (2√21)/3
( 4 - √6 ) / 2√3 = ( 4√3 - √6√3 ) / 23 = (4√3 - 3√2) / 6
If the denominator of the fraction contains addition or subtraction involving a radical expression, to rationalize we need to multiply by the conjugate of the denominator over itself (P - Q)(P + Q) = P² - Q²
2 / (√5 - 1) = (2 / (√5 - 1))*(√5 + 1)/(√5 + 1) = (2(√5 + 1)) / ((√5)² - 1²) = (2(√5 + 1))/(5 - 1) = (2(√5 + 1)) / 4 = (√5 + 1) / 2
8 / (√7 - √3) = 8(√7 + √3) / 4 = 2√7 + 2√3
(4 + 2√5) / (3 + √5) = (4 + 2√5)(3 - √5) / 4 = (2 + 2√5) / 4 = (1 + √5) / 2
If x(2√5 - 3) = 55, find x
x(2√5 - 3) = 55
x = 55 / (2√5 - 3)
x = (55(2√5 + 3)) / 11
x = 5(2√5 + 3)
When a car, initially moving at a speed of v, decelerates with constant acceleration a, its stopping distance d is related to v & a by the formula: v² = 2ad.
If v = 30, and a = 10, find the stopping distance
v² = 2ad 30² = 2*10*d 900 = 2*10*d 90 = 2d 45 = d
Solve v² = 2ad for v
Solve h = (1/2)gt² + vt for g
Solve 1/p + 1/q = 1/f for q
Solving for a variable
v² = 2ad
v = √(2ad)
h = (1/2)gt² + vt 2h = gt² + 2vt 2h - 2vt = gt² g = (2h - 2vt) / t² g = (2h)/t² - (2v)/t
1/p + 1/q = 1/f 1/q = 1/f - 1/p 1/q = p/pf - f/pf 1/q = (p - f)/pf q = pf / (p - f)
v² = 2ad
A/ if v doubles and a stays the same, then d is multiplied by what?
B/ if v triples and a doubles, then d is multiplied by what?
C/ In F = GMP/R², if P triples and R doubles, F is multiplied by what?
D/ In T² = KR^3, if T is multiplied by 5, R is multiplied by what?
Proportional reasoning
a) pick easy nb that satisfy the equation for a start. You can also change any constants to 1
b) change whatever values need to be changed, leave the quantity in the question as an unknown, and solve for it
A/ change the 2 to a 1, and let all three variables equal 1 for start
1² = 111 (remains a valid equation)
double v, and leave d as an unknown
2² = 11d
4 = d
d is multiplied by 4 when v doubles and a stays the same
B/ same start : 1² = 111
make the changes, leaving d as a variable
3² = 12d
d = 9/2
C/ F = GMP/R² change the constants to 1 1 = 1*1*1 / 1² make the changes leaving F as a variable F = 1*1*3 / 2² = 3/4 F is multiplied by 3/4
D/ everything equal to 1 1² = 1*1^3 make the changes 5² = 25 = 1*R^3 R^3 = 25 R = (3)√25 = (3)√(5²) = 5^(2/3)
