5.2a Laws of Logarithms (Example 1)

Posted on April 21, 2020 by user

Example 1

Express the following in term of

$\log_{a} x$ and

$\log_{a} y$ .
(a)

$\log_{a} 3 x$
(b)

$\log_{a} \frac{x}{5}$
(c)

$\log_{a} y^{5}$
(d)

$\log_{a} x y^{3}$
(e)

$\log_{a} \frac{x^{2}}{\sqrt{y}}$
(f)

$\log_{a} \sqrt{\frac{y}{a^{2} x^{3}}}$

Law of Logarithms

5.2a Laws of Logarithms

Posted on April 21, 2020 by user

5.2a Laws of Logarithms

$\begin{array}{l} Law 1: \\ {log}_{a} x y = {log}_{a} x + {log}_{a} y \\ Example: \\ \log_{5} 25 x = {log}_{5} 25 + {log}_{5} x \\ Beware!! \\ {log}_{a} x + {log}_{a} y \neq {log}_{a} (x + y) \end{array}$

$\begin{array}{l} Law 2: \\ {log}_{a} (\frac{x}{y}) = {log}_{a} x - {log}_{a} y \\ Example: \\ \log_{5} \frac{x}{25} = {log}_{5} x - {log}_{5} 25 \\ Beware!! \\ {log}_{a} \frac{x}{y} \neq \frac{{log}_{a} x}{{log}_{a} y} \end{array}$

$\begin{array}{l} Law 3: \\ {log}_{a} x^{m} = m {log}_{a} x \\ Example: \\ \log_{5} y^{5} = 5 {log}_{5} y \\ Beware!! \\ {({log}_{a} x)}^{2} \neq 2 {log}_{a} x \end{array}$

5.2 Logarithms

Posted on April 21, 2020 by user

5.2 Logarithms

$N = a^{x} \Leftrightarrow \log_{a} N = x$

$\log_{a} N = x$ is called the logarithmic form and

$N = a^{x}$ is the index or exponential form.

Note:

The logarithm of a negative number is not defined.
log in the calculator denotes $\log_{10}$ or common logarithm.
$\log_{10}$ may be written as lg.
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 April 21, 2020 by user

5.1 Indices and Laws of Indices (Part 2)

(C) Fractional Indices

$\begin{array}{l} a^{\frac{1}{n}} is a n th root of a . \\ a^{\frac{1}{n}} = \sqrt[n]{a} \\ a^{\frac{m}{n}} is a n th root of a^{m} . \\ a^{\frac{m}{n}} = \sqrt[n]{a^{m}} \end{array}$

Example 1:

Find the value of the followings:

$\begin{array}{l} {(a) 81}^{\frac{1}{2}} \\ {(b) 64}^{\frac{1}{3}} \\ {(c) 625}^{\frac{1}{4}} \end{array}$

Solution:

$\begin{array}{l} {(a) 81}^{\frac{1}{2}} = \sqrt{81} = 9 \\ {(b) 64}^{\frac{1}{3}} = \sqrt[3]{64} = 4 \\ {(c) 625}^{\frac{1}{4}} = \sqrt[4]{625} = 5 \end{array}$

Example 2:

Find the value of the followings:

$\begin{array}{l} (a) 16^{\frac{3}{2}} \\ (b) {(\frac{27}{64})}^{\frac{2}{3}} \end{array}$

Solution:

$\begin{array}{l} (a) 16^{\frac{3}{2}} = {(16^{\frac{1}{2}})}^{3} = 4^{3} = 64 \\ (b) {(\frac{27}{64})}^{\frac{2}{3}} = {(\sqrt[3]{\frac{27}{64}})}^{2} = {(\frac{3}{4})}^{2} = \frac{9}{16} \end{array}$

(D) Laws of Indices

$\begin{array}{l} a^{m} \times a^{n} = a^{m + n} \\ Example: \\ 3^{3} \times 3^{2} \\ = 3^{3 + 2} = 3^{5} = 243 \end{array}$

$\begin{array}{l} a^{m} \div a^{n} = a^{m - n} or \frac{a^{m}}{a^{n}} = a^{m - n}, a \neq 0 \\ Example: \\ 3^{3} \div 3^{2} \\ = 3^{3 - 2} = 3^{1} = 3 \\ or \\ \frac{3^{3}}{3^{2}} = 3^{3 - 2} = 3^{1} = 3 \end{array}$

$\begin{array}{l} {(a^{m})}^{n} = a^{m n} \\ Example: \\ {(7^{3})}^{4} = 7^{3 \times 4} = 7^{12} \end{array}$

$\begin{array}{l} {(a b)}^{n} = a^{n} b^{n} \\ Example: \\ {(15)}^{3} = {(5 \times 3)}^{3} = 2^{3} \times 3^{3} \end{array}$

$\begin{array}{l} {(\frac{a}{b})}^{n} = \frac{a^{n}}{b^{n}}, b \neq 0 \\ Example: \\ {(\frac{3}{5})}^{4} = \frac{3^{4}}{5^{4}} = \frac{81}{625} \end{array}$

Indices and Laws of Indices (Part 1)

Posted on April 21, 2020 by user

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

$\begin{array}{l} (c) {(\frac{2}{7})}^{0} \\ (d) {(- 11 \frac{1}{25})}^{0} \end{array}$

Solution:

(a) 250⁰ = 1

(b) 0.513⁰ = 1

$\begin{array}{l} (c) {(\frac{2}{7})}^{0} = 1 \\ (d) {(- 11 \frac{1}{25})}^{0} = 1 \end{array}$

(B) Negative Integral Indices

$\begin{array}{l} a^{- n} is a reciprocal of a^{n} . \\ a^{- n} = \frac{1}{a^{n}} \end{array}$

Example 2:

Find the value of the followings:

(a) 102 ^-1

(b)–6 ^-3

$\begin{array}{l} (c) {(\frac{1}{3})}^{- 4} \\ (d) {(\frac{2}{5})}^{- 2} \\ (e) - {(\frac{2}{5})}^{- 4} \end{array}$

Solution:

$\begin{array}{l} (a) 102^{- 1} = \frac{1}{102} \\ (b) - 6^{- 3} = \frac{1}{- 6^{3}} = - \frac{1}{216} \\ (c) {(\frac{1}{3})}^{- 4} = {(3)}^{4} = 81 \\ (d) {(\frac{2}{5})}^{- 2} = {(\frac{5}{2})}^{2} = \frac{25}{4} \\ (e) {(- \frac{2}{5})}^{- 4} = {(- \frac{5}{2})}^{4} = \frac{625}{16} \end{array}$

Quadratic Functions, SPM Practice (Long Questions)

Posted on April 18, 2020 by user

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)

$\begin{array}{l} f (x) = 2 x^{2} - p x + q \\ = 2 [x^{2} - \frac{p}{2} x + \frac{q}{2}] \\ = 2 [{(x + \frac{- p}{4})}^{2} - {(\frac{- p}{4})}^{2} + \frac{q}{2}] \\ = 2 [{(x - \frac{p}{4})}^{2} - \frac{p^{2}}{16} + \frac{q}{2}] \\ = 2 {(x - \frac{p}{4})}^{2} - \frac{p^{2}}{8} + q \end{array}$

$\begin{array}{l} ∴ \frac{p}{4} = 1 - - - - - - (1) \\ and - \frac{p^{2}}{8} + q = - 18 - - - (2) \\ From (1), p = 4. \\ Substitute p = 4 into (2) : \\ - \frac{{(4)}^{2}}{8} + q = - 18 \\ - \frac{16}{8} + q = - 18 \\ q = - 18 + 2 \\ = - 16 \end{array}$

(b)

$\begin{array}{l} f (x) = 2 x^{2} - 4 x - 16 \\ f (x) = 0 when it cuts x -axis \\ 2 x^{2} - 4 x - 16 = 0 \\ x^{2} - 2 x - 8 = 0 \\ (x - 4) (x + 2) = 0 \\ x = 4, - 2 \\ Graph f (x) cuts x -axis at x = - 2 and x = 4. \end{array}$

(c)

Quadratic Functions, SPM Practice (Short Questions)

Posted on April 18, 2020 by user

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:

$\begin{array}{l} y = 5 x - 1 ...... (1) \\ y = 2 x^{2} + x + h ...... (2) \\ Substitute (1) into (2), \\ 5 x - 1 = 2 x^{2} + x + h \\ 2 x^{2} + x + h - 5 x + 1 = 0 \\ 2 x^{2} - 4 x + h + 1 = 0 \\ b^{2} - 4 a c < 0 \\ {(- 4)}^{2} - 4 (2) (h + 1) < 0 \\ 16 - 8 h - 8 < 0 \\ 8 < 8 h \\ h > 1 \end{array}$

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

Solution:

$\begin{array}{l} 5 - x - 2 x^{2} \\ = - 2 x^{2} - x + 5 \\ = - 2 [x^{2} + \frac{1}{2} x - \frac{5}{2}] \\ = - 2 [x^{2} + \frac{1}{2} x + {(\frac{1}{4})}^{2} - {(\frac{1}{4})}^{2} - \frac{5}{2}] \\ = - 2 [{(x + \frac{1}{4})}^{2} - \frac{1}{16} - \frac{5}{2}] \\ = - 2 [{(x + \frac{1}{4})}^{2} - \frac{41}{16}] \\ = - 2 {(x + \frac{1}{4})}^{2} + 5 \frac{1}{8} \end{array}$

$\begin{array}{l} 5 - x - 2 x^{2} has a maximum value when \\ 2 {(x + \frac{1}{4})}^{2} = 0 \\ x = - \frac{1}{4} \\ The maximum value of 5 - x - 2 x^{2} is 5 \frac{1}{8} . \end{array}$

Quadratic Functions, SPM Practice (Short Questions)

Posted on April 18, 2020 by user

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).

$\begin{array}{l} f (x) = 2 x^{2} + 6 x + 5 \\ = 2 [x^{2} + 3 x + \frac{5}{2}] \\ = 2 [x^{2} + 3 x + {(3 \times \frac{1}{2})}^{2} - {(3 \times \frac{1}{2})}^{2} + \frac{5}{2}] \end{array}$

$\begin{array}{l} = 2 [{(x + \frac{3}{2})}^{2} - \frac{9}{4} + \frac{5}{2}] \\ = 2 [{(x + \frac{3}{2})}^{2} + \frac{1}{4}] \\ = 2 {(x + \frac{3}{2})}^{2} + \frac{1}{2} \end{array}$

Since a = 2 > 0,

Therefore f (x) has a minimum value when

$x = - \frac{3}{2} .$ . 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 April 18, 2020 by user

(C) Linear Inequality

Example 1
(a) Given

$x = \frac{6 - y}{3}$ , 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 April 18, 2020 by user

Maximum and Minimum Point

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.
The minimum/maximum point can be determined from the equation in this form $f (x) = a (x + p)^{2} + q$ .

Minimum Point

The quadratic function f(x) has a minimum value if a is positive.
The quadratic function f(x) has a minimum value when (x + p) = 0
The minimum value is equal to q.
Hence the minimum point is (-p, q)

Maximum Point

The quadratic function f(x) has a maximum value if a is negative.
The quadratic function f(x) has a maximum value when (x + p) = 0
The maximum value is equal to q.
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)

$\begin{array}{l} f (x) = {(x - 3)}^{2} + 7 \\ a = 1, p = - 3, q = 7 \\ a > 0, the quadratic function has a minimum point \\ Minimum point \\ = (- p, q) \\ = (3, 7) \end{array}$

(b)

$\begin{array}{l} f (x) = - 5 - 3 {(x + 15)}^{2} \\ a = - 3, p = 15, q = - 5 \\ a < 0, the quadratic function has a maximum point \\ Maximum point \\ = (- p, q) \\ = (- 15, - 5) \end{array}$