5.2a Laws of Logarithms

Posted on by

5.2a Laws of Logarithms

  Law 1:   log a x y = log a x + log a y   Example:   log 5 25 x = log 5 25 + log 5 x   Beware!!   log a x + log a y ≠ log a ( x + y )   

  Law 2:   log a ( x y ) = log a x − log a y      Example:   log 5 x 25 = log 5 x − log 5 25   Beware!!   log a x y ≠ log a x log a y   

  Law 3:   log a x m = m log a x   Example:   log 5 y 5 = 5 log 5 y   Beware!!    ( log a x ) 2 ≠ 2 log a x   


5.2 Logarithms

Posted on by

5.2 Logarithms

N = a x           ⇔           log a N = x log a N = x is called the logarithmic form and N = a x is the index or exponential form.



Note:
  1. The logarithm of a negative number is not defined.
  2. log in the calculator denotes log 10 or common logarithm.
  3. log 10 may be written as lg.
  4. If the base is other than 10, the base should be specified, e.g. log 3 81

5.1 Indices and Laws of Indices (Part 2)

Posted on by

5.1 Indices and Laws of Indices (Part 2)

(C) Fractional Indices
   a 1 n  is a  n th root of  a .    a 1 n = a n    a m n  is a  n th root of  a m .    a m n = a m n

Example 1:
Find the value of the followings:
(a) 81 1 2 (b) 64 1 3 (c) 625 1 4

Solution:
(a) 81 1 2 = 81 = 9 (b) 64 1 3 = 64 3 = 4 (c) 625 1 4 = 625 4 = 5

Example 2:
Find the value of the followings:
(a)  16 3 2 (b)  ( 27 64 ) 2 3

Solution:
(a)  16 3 2 = ( 16 1 2 ) 3 = 4 3 = 64 (b)  ( 27 64 ) 2 3 = ( 27 64 3 ) 2 = ( 3 4 ) 2 = 9 16



(D) Laws of Indices

   a m × a n = a m + n   Example:    3 3 × 3 2    = 3 3 + 2 = 3 5 = 243   


   a m ÷ a n = a m − n    or    a m a n = a m − n , a ≠ 0      Example:    3 3 ÷ 3 2    = 3 3 − 2 = 3 1 = 3   or    3 3 3 2 = 3 3 − 2 = 3 1 = 3


   ( a m ) n = a m n   Example:    ( 7 3 ) 4 = 7 3 × 4 = 7 12   


   ( a b ) n = a n b n   Example:    ( 15 ) 3 = ( 5 × 3 ) 3 = 2 3 × 3 3     


   ( a b ) n = a n b n ,   b ≠ 0   Example:    ( 3 5 ) 4 = 3 4 5 4 = 81 625   

Indices and Laws of Indices (Part 1)

Posted on by

Positive Integral Indices
When a real number a is multiplied by itself n times, the result is the nth power of a.

Example:  5×5×5×5 = 54 (5 to the power of 4)

In general, if a is any real number and n is a positive integer, then


The integer n is called the index or exponent and a is the base.



5.1 Indices and Laws of Indices (Part 1)
(A) Zero Indices
The zero index of any number is equal to one.

  a 0 = 1, where a ≠ 0

Example 1:
Find the value of the followings:
(a) 2500
(b) 0.5130
(c)  ( 2 7 ) 0 (d)  ( − 11 1 25 ) 0

Solution:
(a) 2500 = 1
(b) 0.5130 = 1
(c)  ( 2 7 ) 0 = 1 (d)  ( − 11 1 25 ) 0 = 1



(B) Negative Integral Indices

   a −n  is a reciprocal of  a n .      a −n = 1 a n

Example 2:
Find the value of the followings:
(a) 102 -1
(b)  –6 -3
(c)  ( 1 3 ) − 4 (d)  ( 2 5 ) − 2 (e)  − ( 2 5 ) − 4

Solution:
(a)  102 − 1 = 1 102 (b)  − 6 − 3 = 1 − 6 3 = − 1 216 (c)  ( 1 3 ) − 4 = ( 3 ) 4 = 81 (d)  ( 2 5 ) − 2 = ( 5 2 ) 2 = 25 4 (e)  ( − 2 5 ) − 4 = ( − 5 2 ) 4 = 625 16

Quadratic Functions, SPM Practice (Long Questions)

Posted on by

Question 3:
Given that the quadratic function f(x) = 2x2 – px + p has a minimum value of –18 at x = 1.
(a) Find the values of p and q.
(b) With the value of p and q found in (a), find the values of x, where graph f(x) cuts the x-axis.
(c) Hence, sketch the graph of f(x).

Solution:
(a)
f( x )=2 x 2 −px+q =2[ x 2 − p 2 x+ q 2 ] =2[ ( x+ −p 4 ) 2 − ( −p 4 ) 2 + q 2 ] =2[ ( x− p 4 ) 2 − p 2 16 + q 2 ] =2 ( x− p 4 ) 2 − p 2 8 +q


∴ p 4 =1−−−−−−( 1 ) and − p 2 8 +q=−18−−−( 2 ) From( 1 ),p=4. Substitute p=4 into ( 2 ): − ( 4 ) 2 8 +q=−18   − 16 8 +q=−18  q=−18+2    =−16


(b)
f( x )=2 x 2 −4x−16 f( x )=0 when it cuts x-axis 2 x 2 −4x−16=0 x 2 −2x−8=0 ( x−4 )( x+2 )=0 x=4,−2 Graph f( x ) cuts x-axis at x=−2 and x=4.

(c)

Quadratic Functions, SPM Practice (Short Questions)

Posted on by

Question 3:
The straight line y = 5x – 1 does not intersect with the curve y = 2x2 + x + h.
Find the range of values of h.

Solution:
y=5x−1         ...... (1) y=2 x 2 +x+h ...... (2) Substitute (1) into (2), 5x−1=2 x 2 +x+h 2 x 2 +x+h−5x+1=0 2 x 2 −4x+h+1=0                   b 2 −4ac<0 ( −4 ) 2 −4( 2 )( h+1 )<0              16−8h−8<0                             8<8h                             h>1

Question 4:
Find the maximum value of the function 5 – x – 2x2 , and the corresponding value of x.

Solution:
5−x−2 x 2 =−2 x 2 −x+5 =−2[ x 2 + 1 2 x− 5 2 ] =−2[ x 2 + 1 2 x+ ( 1 4 ) 2 − ( 1 4 ) 2 − 5 2 ] =−2[ ( x+ 1 4 ) 2 − 1 16 − 5 2 ] =−2[ ( x+ 1 4 ) 2 − 41 16 ] =−2 ( x+ 1 4 ) 2 +5 1 8


5−x−2 x 2  has a maximum value when 2 ( x+ 1 4 ) 2 =0               x=− 1 4 The maximum value of 5−x−2 x 2  is 5 1 8 .

Quadratic Functions, SPM Practice (Short Questions)

Posted on by

Question 1:

Find the minimum value of the function f (x) = 2x2 + 6x + 5. State the value of xthat makes f (x) a minimum value.

Solution:

By completing the square for f (x) in the form of f (x) = a(x + p)2 + q to find the minimum value of f (x).

f ( x ) = 2 x 2 + 6 x + 5 = 2 [ x 2 + 3 x + 5 2 ] = 2 [ x 2 + 3 x + ( 3 × 1 2 ) 2 − ( 3 × 1 2 ) 2 + 5 2 ]
= 2 [ ( x + 3 2 ) 2 − 9 4 + 5 2 ] = 2 [ ( x + 3 2 ) 2 + 1 4 ] = 2 ( x + 3 2 ) 2 + 1 2

Since a = 2 > 0,
Therefore f (x) has a minimum value when x = − 3 2 . . The minimum value of f (x) = ½. 

Question 2:

The quadratic function f (x) = –x2 + 4x + k2, where k is a constant, has a maximum value of 8.
Find the possible values of k.

Solution:
f (x) = –x2 + 4x + k2
f (x) = –(x2 – 4x) + k2 ← [completing the square for f (x) in
the form of f (x) = a(x + p)2+ q]
f (x) = –[x2 – 4x + (–2)2 – (–2)2] + k2
f (x) = –[(x – 2)2 – 4] + k2
f (x) = –(x – 2)2 + 4 + k2

Given the maximum value is 8.
Therefore, 4 + k2 = 8
k2 = 4
k = ±2

 

 

 

 

3.4 Quadratic Inequalities (Part 2)

Posted on by
(C) Linear Inequality 

Example 1
(a)  Given x = 6 − y 3  , find the range of values of x for which y > 9 .
(b) Given 2 x + 3 y − 6 = 0  , find the range of values of x for which y < 4 .






(D) Quadratic Inequalities 

Example 2
Find the range of values of x which satisfy  the following inequalities:
(a) (2x + 1) (3x – 1) < 14
(b) (x– 2) (5x – 4) + 1 > 0

3.2 Maximum and Minimum Value of Quadratic Functions

Posted on by

Maximum and Minimum Point

  1. A quadratic functions f ( x ) = a x 2 + b x + c can be expressed in the form f ( x ) = a ( x + p ) 2 + q by the method of completing the square.
  2. The minimum/maximum point can be determined from the equation in this form f ( x ) = a ( x + p ) 2 + q .
Minimum Point
  1. The quadratic function f(x) has a minimum value if a is positive. 
  2. The quadratic function f(x) has a minimum value when (x + p) = 0
  3. The minimum value is equal to q.
  4. Hence the minimum point is (-p, q)

Maximum Point

  1. The quadratic function f(x) has a maximum value if a is negative.
  2. The quadratic function f(x) has a maximum value when (x + p) = 0
  3. The maximum value is equal to q.
  4. Hence the maximum point is (-p, q)



Example
Find the maximum or minimum point of the following quadratic equations
a. f ( x ) = ( x − 3 ) 2 + 7
b. f ( x ) = − 5 − 3 ( x + 15 ) 2

Answer:
(a)
f ( x ) = ( x − 3 ) 2 + 7 a = 1 , p = − 3 , q = 7 a > 0 ,  the quadratic function has a minimum point Minimum point = ( − p , q ) = ( 3 , 7 )

(b)
f ( x ) = − 5 − 3 ( x + 15 ) 2 a = − 3 ,   p = 15 ,   q = − 5 a < 0 ,  the quadratic function has a maximum point Maximum point = ( − p , q ) = ( − 15 , − 5 )