Sunday 30 June 2013

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

Saturday 29 June 2013

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    






Tuesday 25 June 2013

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

-6x24
-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

Monday 24 June 2013

LOGARITHMS - 2



If Logb125 – log 0.00001 = 8; find b

Solution

Logb125 – log 0.00001 = 8;            [0.00001=10-5 ]

Logb125 – log 10-5 = 8      

Logb125 – (-5log 10) = 8

Logb125 + 5log 10 = 8   [remember log 10 = 1]

Logb125 + (5 x 1) = 8

Logb125 + 5 = 8

Logb125 = 8 - 5

Logb125 = 3     [Change it into exponential notation]

125 = b3

53 = b3

b = 5


Hence b = 5       


Friday 21 June 2013

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 Logy81 – log53125 = -1; find y

Solution

Logy81 – Log555 = -1        3135=5x5x5x5x5=55 by prime factorization

Logy81 – 5Log55 = -1

Logy81 – 5x1 = -1

Logy81 – 5 = -1

Logy81 = -1 + 5

Logy81 = 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  104 + -70


4.02543 x  10-66 answer    




Wednesday 19 June 2013

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




Thursday 13 June 2013

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   
 
 



Wednesday 12 June 2013

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 = ?

An = 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      
 
 
 




Tuesday 11 June 2013

FUNCTIONS - 4


If f(x)=20x, find f-1(1/400)

Solution

f(x)=20x

let f(x) = y

then y=20x

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