Monday, 17 July 2017

LOGARITHMS 22


If logx 2401 – log3729 = -4; find x

solution

logx 2401 – log3729 = -2

logx 2401 – log336 = -2

logx 2401 – 6log33 = -2

logx 2401 – 6 x 1 = -2

logx 2401 – 6 = -2

logx 2401 = -2 + 6

logx 2401 = 4 

2401 = x4        2401=7x7x7x7=74 by prime factors

74 = x4      Powers cancel out

x = 7


Hence x = 7.

TRY THIS………….


If logm2048 – log6216 = 8; find m

EXPONENTIALS 10


Simplify 44
               8-2/3

Solution

44
    8-2/3

= 44 x 1
           8-2/3

=  44 x 82/3        [Since 1/a-n = an]

=  44 x (81/3)2        [Since 81/3 = cube root of 8 = 2]

=  44 x (2)2        

=  44 x 4

= 176       
          
Hence   44        = 176
             8-2/3


TRY THIS…..  

simplify 20
             125-2/3



ARITH. PROGRESSION 3


The 1st term of an A.P. is 80 and the common difference is 27. Find the nth term.

Solution

A1= 80, d = 27

An = A1 + (n-1)d

An = 80 + (n-1)27

An = 80 + 27n – 27

An = 27n - 53



Hence the nth term is 27n - 53.



TRY THIS……………


The 1st term of an A.P. is 24 and the common difference is 34. Find the nth term.

LOGARITHMS 21

If log 2= 0.3010; find the value of log 1250 without using tables.

Solution

=log1250

=log(10,000 ÷8)

=log10,000 –log8

= log104 - log23

= 4log10 - 3log2

=(4 x 1) - (3 x 0.3010)

= 4  -  0.9030

=3.097

Hence log 1250 =3.097

TRY THIS……………


If log 2= 0.3010; find the value of log 2500 without using tables.

ALGEBRA 12


If 7x + 5y - 56 = 0; find x-intercept.

Solution

x-intercept is when y=0.

7x + 5y -56 = 0

7x + 5(0) - 56 = 0

7x - 56 = 0

7x = 0+ 56

7x = 56
7      7

x= 8

Hence x-intercept= 8

TRY THIS………..


If 8x - 7y - 40 = 0; find x-intercept.

Sunday, 16 July 2017

EXPONENTIALS 9


If 2(6m-7) x 3(n+8) = (344) x (265); Find m and n.

solution

equating equal bases;

2(6m-7) =  265

2(6m-7)265

6m - 7 = 65

6m  = 65 + 7

6m  = 72

16m  =  725
16         61

m =  12

again for n we equate equal bases.

3(n+8) = 344

3(n+8) = 344

n + 8 = 44

n = 44 - 8

n = 36

Hence m = 12 and n = 36


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



If 8(2m-17) x 11(5n+40) = (1170) x (843); Find m and n.

GEOM. PROGRESSION 3


Find the sum of the 1st nine terms of the geometrical progression 3+6+12+24+…….

Solution

G1 = 3, r=2, n=9

Sn = G1(rn-1)/r-1
               
S9 = 3(29-1)/2-1
             
S9 = 3(29-1)/1
              
S9 = 3(29-1)

S9 = 3(512-1)

S9 = 3(511)

S9 = 1533


Hence the sum of the 1st nine terms is 1533.

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


Find the sum of the 1st seven terms of the geometrical progression 5+10+20+40+…….

ALGEBRA 11


48 + 24 =56
        a

Solution

48 + 24 =56
        a

   24  = 56 - 48
   a

   24  =  8
    a

1 a  x  24  =  8 x a
        1 a

24 = 8a

24  = 8a
8       8

3 = a

Hence a=3.

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

If 30 + 34 = 47; find m

           m

SETS 4


If n(A)= 85 , n(B)= 96 and n(AuB)= 130, find n(AnB).

Solution

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

130 = 85 + 96 - n(AnB)

130 = 181 - n(AnB)

130 - 181 = - n(AnB)

-51 = -n(AnB)

n(AnB) = 51 [after dividing by -1 both sides]

Hence n(AnB) = 51 answer

TRY THIS………….


If n(A)= 92 , n(B)= 95 and n(AuB)= 137, find n(AnB).

GEOM. PROGRESSION 2


The sum of the 1st five terms of a geometrical progression is 484. If the common ratio is 3 find the 4th term.

Solution

S5 = 484, G1 = ?, r=3, n=5, A4=?

We use the summation formula to find the 1st term.
Sn = G1(rn-1)/r-1
               

S5 = G1(r5-1)/r-1
               

484= G1(35-1)/3-1
               

484= G1(243-1)
                  2

2 x 484= G1(242)
                  
968= 242 G1

4 968   =  G1
  242

G1 = 4

Now we solve for the 4th term.

Gn = G1rn-1

G4 = G1r4-1

G4 = G1r3

G4 = 4 x (3 x 3 x 3)

     = 108


Hence the 4th term is 108.

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


The sum of the 1st five terms of a geometrical progression is 484. If the common ratio is 3 find the 7th term.