ALGEBRA 1B

Compute the value of x if x³ - 25 = 103 - x³.

Solution

x³ - 25 = 103 - x³.

x³ + x³ - 25 = 103

2x³ - 25 = 103

2x³ = 103 + 25

2x³ = 128

2x³ = 128⁶⁴

2        2

x³ = 64

x= 4  [since the cube root of 64 is 4]

Hence x=4

TRY THIS……………………… 

Compute the value of x if 2x³ - 101 = 74 - x³.
Answer: x=5.

LOGARITHMS 1C

If log₃(10x + 7)=3; find x.

Solution

log₃(10x + 7)=3

 (10x + 7)=3³         

                                            

 10x + 7=27

10x = 27 – 7

10x =  20

10x =  20²   

10      10

x = 2

Hence x = 2

TRY THIS…………….

If log₃(5x + 1)=4; find x

COORDINATE GEOMETRY 1A

If x-15⁰ and 4x + 35⁰ are complementary angles, find the value of x.

solution

Complementary angles add up to 90⁰.

so,  x-15⁰ + 4x + 35⁰ = 90⁰.

 x + 4x + 35⁰ - 15⁰ = 90⁰.

 5x + 20⁰ = 90⁰.

  5x = 90⁰ -  20⁰

  5x = 70⁰

  5x = 70¹⁴

  5       5

Hence x = 14

TRY THIS...............

 If 3x - 22⁰ and 2x + 87⁰ are complementary angles, find the value of x.

SLOPE OR GRADIENT 1A

Find the slope of a line which passes through (-3, -4) and (8,-11)

Solution

x₁ = -3,  y₁ =-4,  x₂ = 8,  y₂ = -11

m = y2 –y1

      x2 – x1

m =   -11 –(-4)

          8 –(-3)

m =   -11 + 4

          8 + 3

m =    - 7

          11

Hence the slope is -7/11

TRY THIS...................................

Find the slope of a line which passes through (-11, -4) and (4,-15)

POLYNOMIALS 1A

If f(x) = x⁴ + kx² + 6x + 7 has a remainder of 22 when divided by x+2; find k.

solution

f(x) = x⁴ + kx² + 6x + 7

x + 2 = 0

x = -2

f(x) = (-2)⁴ + k(-2)² + 6(-2) + 7 = 22

16 + 4k + (-12) + 7 = 22

16 + 4k - 12 + 7 = 22

16 + 4k - 5 = 22

4k + 16 - 5 = 22

4k + 11= 22

4k = 22 - 11

4k = 11

4k = 11

4      4

k = ¹¹/4

hence k=¹¹/4

TRY THIS......................

 If f(x) = x⁴ - kx² + 3x - 11 has a remainder of 16 when divided by x-3; find k.

LOGARITHMS 1B

Evaluate Log100000 +  log 0.00001 + log₃243

Solution

= Log100000 +  log 0.001 + log₃243

= Log10⁵ +  log 10^-5 + log₃3⁵

= 5Log10 +  (-5log 10) + (5log₃3)    [since log_aaⁿ  = nlog_aa]

= (5x1) + (-5x1) + (5x1)            [since log_aa = 1]

= 5 + (-5) + (5)

= 5

Hence Log100,000 +  log 0.00001 + log₃243 = 5

TRY THIS...............................

Evaluate Log10,000,000 +  log 0.01 + log₃729

ARITHMETICS 1B

Evaluate 1168² – 832²

Solution

We apply difference of  two squares: a² - b² = (a-b)(a+b).

1168² – 832² = (1168 + 832)( 1168 - 832)

                     = (2000)( 336)

                     = 672000   [only multiply 336 and 2, then add three zeros]

Hence 1168² – 832² = 672000

TRY THIS……………………

Evaluate 1253² – 747²

ALGEBRA 1A

Expand 8w(5w – 7)

solution

= 8w(5w – 7)

= (8w x 5w) – (8w x 7)

= 40w² – 56w answer

TRY THIS………..

Expand 2a(7a+ 30)

EXPONENTIALS 1A

If 5^2w (40^w) = 1000000 ; Find w.

Solution

5^2w (40^w) = 1000000

(5²)^w (40^w) = 10⁶

(25)^w (40^w) = 10⁶

(25 x 40)^w = 10⁶

(1000)^w = 10⁶

(10³)^w = 10⁶

10^3w = 10⁶   (Bases are alike, so they cancel out)

3w = 6

3w = 6²

3       3

w = 2

TRY THIS…………………………….

If 4^2t (4^t) = 128 ; Find t.

LOGARITHMS 1A

Change the following into Logarithmic form.

i)  4² = 16

ii)  5³ = 125

iii)  3⁰ = 1

Solution

i) 4² = 16 ⇔ log₄ 16 = 2

ii) 5³ = 125 ⇔ log₅ 125 = 3

iii) 3⁰ = 1 ⇔ log₃ 1 = 0

TRY THIS………………….

Change the following into Logarithmic form.

i)  7² = 49
ii)  3⁴ = 81
iii)  8⁰ = 1