LOGARITHMS - 4

Evaluate Log100000 +  log 0.01 + log₃243

Solution

= Log100000 +  log 0.01 + log₃243

= Log10⁵ +  log 10^-2 + log₃3^-5

= 5Log10 +  (-2log 10) + (-5log₃3)

= (5x1) + (-2x1) + (-5x1)

= 5 + (-2) + (-5)

= -2

Hence Log100000 +  log 0.01 + log₃243 = -2

LOGARITHMS - 3

Evaluate Log10000000 +  log 0.001

Solution

= Log10000000 +  log 0.001

= Log10⁷ +  log 10^-3

= 7Log10 +  (-3log 10)

= (7x1) + (-3x1)

=7 + (-3)

= 4

Hence Log10000000 +  log 0.001 = 4

INEQUALITIES - 2

Find x in the following inequality 2x+50 ≤ 72

Solution

2x+50 ≤ 72

2x ≤ 72 – 50

2x≤ 22

2x  ≤  22

2         2

x ≤ 11

Hence x ≤ 11

ABSOLUTE VALUE EQUATIONS

If │4x – 7 │= 17; Find  x

Solution

±(4x – 7 )= 17

4x -7= 17    OR   –(4x-7) = 17

4x – 7 = 17    OR   -4x+7=17

4x = 17 + 7    OR   -4x= 17-7

4x = 24     OR   -4x  =  10

4      4                   4        4

x = 6  OR  x = 5/2

Hence x = 6 OR x = 5/2    

INEQUALITIES - 1

Find x in the following inequality. 48 – 6x ≤ 72

Solution

48 – 6x ≤ 72

48-48 – 6x ≤72-48    subtracting 48 on both sides.

0 – 6x ≤ 24

-6x ≤ 24

-6x ≥ 24

-6      -6

x  ≥ -4  The inequality sign changes because x has changed from negative to positive.

Hence x  ≥ -4

ALGEBRA - 3

The sum of four consecutive numbers is 362. Find the 3rd number.

Solution

Let the numbers be as shown in the table below

1st number	2nd number	3rd number	4th number	TOTAL
n	n+1	n+2	n+3	362

Then, n + (n+1) + (n+2) + (n+3) = 362

4n + 1+2+3= 362

4n + 6 = 362

4n = 362 – 6

4n = 356 

4n = 356

 4        4

n = 89

3rd number = n +2

                       = 89 + 2

                      = 91

Hence the 3rd number is 91

LOGARITHMS - 2

If Log_b125 – log 0.00001 = 8; find b

Solution

Log_b125 – log 0.00001 = 8;            [0.00001=10^-5 ]

Log_b125 – log 10^-5 = 8      

Log_b125 – (-5log 10) = 8

Log_b125 + 5log 10 = 8   [remember log 10 = 1]

Log_b125 + (5 x 1) = 8

Log_b125 + 5 = 8

Log_b125 = 8 - 5

Log_b125 = 3     [Change it into exponential notation]

125 = b³

5³ = b³

b = 5

Hence b = 5       

CO-ORDINATE GEOMETRY - 3

A line passes through (2a, 7) and (8,5a). If its slope is 4, find a.

Solution

x1=2a, y1 = 7, x2 = 8, y2 = 5a

Slope(m) =    y2 – y1

                     x2 – x1

               4 =  5a – 7

                      8 – 2a

               4 =    5a – 7     [cross multiplying]

               1       8 – 2a

4(8 – 2a) = 5a – 7

32 – 8a = 5a – 7

32 – 8a + 7 = 5a

32 + 7 = 5a+ 8a

39 = 13a

39 = 13a

13     13

3 = a

Hence a = 3

LOGARITHMS - 1

If Log_y81 – log₅3125 = -1; find y

Solution

Log_y81 – Log₅5⁵ = -1        3135=5x5x5x5x5=5⁵ by prime factorization

Log_y81 – 5Log₅5 = -1

Log_y81 – 5x1 = -1

Log_y81 – 5 = -1

Log_y81 = -1 + 5

Log_y81 = 4

81 = y⁴                   81=3x3x3x3 = 3⁴ by prime factorization

3⁴ = y⁴

3 =y

Hence y = 3      

EXPONENTS - 2

Evaluate (789.3 x 10^-20) x (51 x 10^-50) giving your answer in standard form.

Solution

(789.3 x 51) x (10^-20x 10^-50)  Group decimals alone, and powers alone

= 40254.3 x  10^{-20 + -50}

= 40254.3 x  10^-70

= 4.02543 x  10⁴ x 10^-70

= 4.02543 x  10^{4 + -70}

4.02543 x  10^-66 answer    

QUADRATIC - 2

If 4x² – 20x + t is a perfect square, find t

Solution

a = 4, b = -20, c = t.

b² = 4ac

(-20)² = 4 x 4 x t

400 = 16t

400 = 16t

16      16
           

t = 25

Hence t = 25

CIRCLE - 2

Find the value of y.

solution

From circle geometry,

(QR)(QS) = (PQ)²

y(y + 9) = 6²

y² + 9y = 36

y² + 9y – 36 = 0

By factorization

y² + 12y – 3y – 36 = 0

(y² + 12y) – (3y – 36) = 0

y(y + 12) – 3(y + 12) = 0

(y - 3 )( y + 12) = 0                             

y - 3 = 0  OR  y + 12 = 0

y = 3 OR y = -12 

Since we don’t have negative length, we take y=3cm

SEQUENCE AND SERIES - 3

The 1st term of arithmetic progression is 7 and the common difference is 40. Find the nth term

solution

A1=7, d= 40, n=?

An =A1 + (n-1) d

An=7 + (n-1)40

An=7 + 40n -40

An=40n + 7-40

An=40n – 33

Hence the nth term is An=40n – 33   

 

 

SEQUENCE & SERIES - 2

The 1st term of arithmetic progression is 12 and the common difference is 30. Find the 7th term

solution

A₁  = 12, d = 30, n = 7, A7 = ?

A_n = A₁ + (n-1)d

A₇  = A₁ + (7-1)d

A₇  = A₁ + 6d

A₇  = 12 + (6x30)

A₇  = 12+180

A₇  = 192

Hence the 7th term is 192

AREA - 1

Area of a rectangle is 77 cm². Find its perimeter if its width is 7cm.

 

Solution

 

First we find length

 

A = l x w

 

77 = l x 7

 

77 = 7l

7       7

 

11 = l

 

Now length is 11cm.

 

We can find perimeter since we have both length and width.

 

Let l = length and w = width;

 

P = 2(l + w)

 

   = 2(11 + 7)

 

   = 2 x 18

 

   = 36

 

Hence perimeter is 36cm      

 

 

 

FUNCTIONS - 4

If f(x)=20^x, find f^-1(¹/₄₀₀)

Solution

f(x)=20^x

let f(x) = y

then y=20^x

we change it into logarithmic form, and it will be,

log₂₀y=x;

x= log₂₀y after re-arranging,

Then we interchange x and y variables,

y^-1 = log₂₀x

f^-1(x) = log₂₀x

f^-1(¹/₄₀₀) = log₂₀(¹/₄₀₀)

              =  log₂₀(400^-1)

  

              = -1 x log₂₀(400); 400 is a perfect square giving (20 x 20=20²)

              = -1 x log₂₀(20)²

   

              = -1 x 2 x log₂₀20

  

              = -1 x 2 x 1

              = -2

Hence f^-1(¹/400) = -2