Friday, 1 June 2018

LOG 2


If log9(32x + 17)=2; find x.

   Solution

Log9(32x + 17)=2

(32x + 17)=92       
                                            
32x + 17=81

32x =81 – 17

32x = 64

32x =   64   
32        32

x = 2

Hence x = 2

TRY THIS………………


If log7(2t - 65)=2; find t

VARIATIONS 1


x is inversely proportional to y. x=12 while y=20. Find y when x is 4.

Solution

x k
      y

x = k
      y

12 = k
        20

k = 20 x 12

k = 240

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

k = 240, y= ?, x = 4.

x = k
      y

4 = 240
        y

y x 4 = 240  x  y   (multiply by y on both sides)
              y


4y = 240
4      4

y = 60.


TRY THIS…………….


x is inversely proportional to y. x=200 while y=4. Find y when x is 80.

LOG 1


Evaluate Log60 – Log0.3 + Log5000.

Solution

= Log60 – Log0.3 + Log5000.

= Log(60 x 5000)
               0.3

= Log(300000)
             0.3

= Log(3,000,000)
              3


= Log 1,000,000
             

= Log101,000,000

= Log10106
            
= 6Log1010

= 6 x 1

= 6

Log60 – Log0.3 + Log5000 = 4

TRY THIS……………………….


Evaluate Log6000 – Log4.8 + Log80,000

Friday, 4 May 2018

SETS 2


If n(AnB)=45, n(B)= 96 and n(AuB)= 146, find n(A)

Solution

n(AuB) = n(A) + n(B) - n(AnB)

146 = n(A)  + 96 - 45

146 = n(A)  +  51

146 - 51 = n(A) 

95 = n(A) 

Hence n(A)  = 95 answer


TRY THIS………….



If n(AnB)=44, n(B)= 97 and n(AuB)= 145, find n(A)


FUNCTIONS 2


Given that F(x) = 3x  +  11. Find F(-4)

Solution

F(x) = 3x  +  11

F(-4) = 3(-4)  +  11

F(-4) = -12  +  11

F(-4) = -1


Hence  F(-4) = -1

TRY THIS………………..



Given that F(x) = 14x  +  38. Find F(-4)

LOGARITHMS 2




Simplify Log2256 - Log3243

Solution

= Log21024 - Log3243

= Log2210 - Log335    [Since 256=28 and 243=35].

= 10Log22 - 5Log33    [since Logaa = 1]

= (10 x 1) - (5 x 1)

= 10 - 5

= 5

Hence Log21024 - Log3243 = 5

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

Simplify Log22048 – Log5625

Thursday, 3 May 2018

PROBABILITY 2


A card is chosen at random from a deck of 52 cards. It is then replaced and a second card is chosen. What is the probability of choosing a jack and then an eight?

solution
P(jack)
 = 
 4 
52
P(8)
 = 
 4 
52
P(jack and 8)
 = 
P(jack)
 · 
P(8)

 = 
 4 
 · 
 4 
52
52
 = 
  16  
2704
 = 
  1  
169


Hence probability of choosing a jack and then an eight is 1/69


TRY THIS………….


A card is chosen at random from a deck of 52 cards. It is then replaced and a second card is chosen. What is the probability of choosing a king and then a three?

FUNCTIONS 1


If F(x) = 2x + 20; Find F-1(x).

Solution

HINT: F-1(x) means inverse.

PROCEDURE:
Make x the subject and then interchange x and y variables.

Let y=F(x)

So,  y= 2x + 20

y – 20 = 2x

y – 20  = 2x
   2          2

y – 20   = x
   2

x  =   y – 20     after rearranging
           2

 y-1 = x – 20     after interchanging x and y variables.
             2

F-1(x) = x – 20     after interchanging x and y variables.
                2

Hence, F-1(x) = x – 20     
                             2


TRY THIS……………………………


If F(x) = 2x - 27; Find F-1(x).

LOGARITHMS 1




Evaluate Log100,000 -  log 0.001 -  log 0.0001

Solution

= Log100,000  -  log 0.001 -  log 0.0001

= Log105 -  log 10-3 – log 10-4

= 5Log10 - (-3 log 10) – (-4log 10)

= (5x1) - (-3x1) - (-8x1)

=5 - (-3) – (-4)

= 5 + 3 + 4

=12
  
Hence Log100,000 -  log 0.001-  log 0.0001 = 12


TRY THIS……………………… 


Evaluate Log100,000 -  log 0.000001 -  log 0.00000001

Wednesday, 2 May 2018

SUM A.P. 1

The first term of an AP is 41 and the last term is 107.If the arithmetic progression consists of 30 terms, calculate the sum of all the terms. 

Solution

A1 = 41, n=30, An = 107, S30 =?

Sn = n(A1 + An)
       2

S30 = 30(41 + 107)
         2

S30 = 15 x 148

S30 = 2220



Hence the sum of all 20 terms is 1480.


TRYTHIS………………….


The first term of an AP is 15 and the last term is 113. If the arithmetic progression consists of 30 terms, calculate the sum of all the terms.