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SS1: MATHEMATICS - 1ST TERM

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  1. Number Base System I | Week 1
    6Topics
    |
    2 Quizzes
  2. Number Base System II | Week 2
    3Topics
    |
    1 Quiz
  3. Number Base System III | Week 3
    2Topics
    |
    1 Quiz
  4. Modular Arithmetic I | Week 4
    2Topics
  5. Modular Arithmetic II | Week 5
    2Topics
  6. Modular Arithmetic III | Week 6
    3Topics
    |
    1 Quiz
  7. Indices I | Week 7
    2Topics
  8. Indices II | Week 8
    1Topic
    |
    1 Quiz
  9. Logarithms I | Week 9
    3Topics
  10. Logarithms II
    1Topic
    |
    1 Quiz
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Evaluate the following: (Try to work these examples out on your own using the laws of indices and then check the solutions by clicking ‘view solutions’ )

(a) \( \scriptsize 5x^{-2}y \:\times\: 2x ^{-5}y^2\)

(b) \( \scriptsize 2^{4}\: \times \: \normalsize \left(\frac {1}{8}\right)^{^{-1} } \scriptsize \: \times \: 4^0\)

(c) \( \scriptsize \left[ 8^{-1}\normalsize \left(\frac{ry^2}{4ry^3}\right )^{^0}\scriptsize \: \times \: 8x \right ]^{-2}\)

(d) \( \frac {1}{81}\scriptsize (3^3 \: – \: 3^{x\: -\:1})\)

(e) \( \scriptsize \sqrt {x^2 \: +\: 2xy\: + \:y^2}\)

(f) \( \frac {1}{a^{4x}} \scriptsize \left[a^{4x}\: – \: (a^{-2x})^0 \right]\)

(g) \( \left( \frac{9a^2}{25} \right)^{^{-\frac{1}{2}}}\)

(h) \( \frac {-x^3}{(-x^6)( -x)}\scriptsize \: \times \: x^4 \)

(i) \(\left (\frac {8y^3}{(27p^6)}\right)^{^{- \frac {2}{3}}}\)

(j) \( \scriptsize 32 ^{^{ -\normalsize\:0.4}}\)

(k) \( \left(\frac{1}{8}\right) ^{^{ – 1 \frac{1}{3}}}\)

(l) \( \left(\frac {9}{4} \scriptsize a^6 \right )^{ ^{-\normalsize 1.5}}\)

SOLUTION (a)

\( \scriptsize 5x^{-2}y \times 2x ^{-5}y^2\)

Group like terms together

\( \scriptsize 5x^{-2} \times 2x ^{-5} \times y \times y^2\)

 

From the Laws of Indicies \( \scriptsize X^a \times X^b = X^{a + b} \)

∴ \( \scriptsize 10x^{-2 + (-5)} \times y^{1 + 2}\)

= \( \scriptsize 10x^{-7} \times y^{3}\)

From the Laws of Indicies \( \scriptsize X^{-a} = \normalsize \frac {1}{X^a} \)

∴ \( \scriptsize x^{-7} = \normalsize \frac{1}{x^{7}} \)

∴ \( \scriptsize 10\: \times \: \normalsize \frac{1}{x^{7}} \scriptsize \: \times \: y^{3}\\ = \normalsize \frac{10y^3}{x^{7}} \)

SOLUTION (b)

\( \scriptsize 2^{4} \: \times \: \normalsize\left (\frac {1}{8}\right )^{^{-1}} \scriptsize \: \times \:  4^0\)

From the Laws of Indicies \( \scriptsize X^0 = 1 \scriptsize \; and \; X^{-a} = \normalsize \frac {1}{X^a} \)

∴\( \scriptsize 4^0 = 1 \: and \: \normalsize \left( \frac {1}{8}\right)^{^{-1}} = \left(\frac {8}{1}\right)^{^{1}} \scriptsize = 8 \)

∴ \( \scriptsize 2^{4} \times \normalsize \left(\frac {1}{8}\right)^{^{-1}} \scriptsize \times 4^0\)

= \( \scriptsize 2^{4} \times \scriptsize 8 \times 1 \)

Note: 8 = 2 x 2 x 2 = \( \scriptsize 2^3 \)

∴ = \( \scriptsize 2^{4} \times 2^{3} \)

From the Laws of Indicies \( \scriptsize X^a \:\times \: X^b = X^{a + b} \)

∴ = \( \scriptsize 2^{4+ 3} = 2^7 \: or \: 128 \)

SOLUTION (c)

\( \scriptsize \left[ 8^{-1}\normalsize \left(\frac{ry^2}{4ry^3}\right )^{^0}\scriptsize \: \times \: 8x \right ]^{-2}\)

From the Laws of Indicies \( \scriptsize X^0 = 1 \)

∴ \( \left (\frac{ry^2}{4ry^3}\right )^{^0} \scriptsize = 1 \)

∴ \( \scriptsize \left [8^{-1}\: \times \: 1 \scriptsize \: \times \:  8x\right ]^{-2}\)

= \( \left [ \normalsize \frac{1}{8} \scriptsize \: \times \:  8x\right ]^{-2}\)

= \( \scriptsize x^{-2} \\ = \normalsize \frac{1}{x^2}\)

SOLUTION (d)

\( \frac {1}{81}\scriptsize (3^3 – 3^{x-1})\)

We know that \( \frac {1}{81}= \frac {1}{3^4} \)

∴\( \frac {1}{3^4}\scriptsize (3^3\: – \:3^{x-1})\)

=\( \frac {3^3}{3^4} – \frac {3^{x-1}}{3^4}\)

From the Laws of Indicies = \( \scriptsize X^a \div X^b \: or \: \normalsize \frac{X^a}{X^b} \scriptsize= X^{a\: – \:b} \)

=\( \scriptsize 3^{(3 \:- \:4)}\: – \: 3^{(x\: -\:1 \:- \:4)} \)

=\( \scriptsize 3^{-1}\: – \:3^{(x \:-\:5)} \)

By Factorisation

Remember: \( \normalsize \frac{3^{-1}}{3^{-1}} \scriptsize = 1 \: and \: \normalsize \frac{3^{(x \: – \: 5)}}{3^{-1}} \scriptsize = 3^{x \: – \: 4} \)

= \( \scriptsize 3^{-1} ( 1 \: – \:3^{x\:-\:4}) \)

or \( \frac {1\: – \:3^{x\: -\:4}} {3}\)

SOLUTION (e)

\( \scriptsize \sqrt {x^2 \:+ \:2xy \:+ \:y^2}\)

recall the complete square of \( \scriptsize x^2 \: +\: 2xy \:+ \:y^2 \:is \: (x \:+ \:y)^2 \)

∴\( \scriptsize \sqrt {x^2\: +\: 2xy \:+ \:y^2} = \sqrt{ (x \:+ \:y)^2} \)

= \(\scriptsize\left [(x \:+ \:y)^2 \right ]^{^{\normalsize \frac{1}{2}}}\)

= \(\scriptsize (x \:+ \:y)^{2 \: \times \: \normalsize \frac{1}{2}}\)

= \(\scriptsize (x \:+ \:y)^1\)

= \(\scriptsize (x \:+\: y)\)

SOLUTION (f)

\( \frac {1}{a^{4x}} \scriptsize \left [a^{4x} \: -\: (a^{-2x})^0\right ]\)

 

From the Laws of Indices

\(\scriptsize X^0 = 1 \: \therefore \:  (a^{-2x})^0 = 1 \)

= \( \frac {1}{a^{4x}} \scriptsize \left (a^{4x} \: – \:  1 \right )\)

Open the bracket

= \( \frac {a^{4x}}{a^{4x}} \: – \:\frac {1}{a^{4x}} \)

= \( \scriptsize 1 \: – \:  \normalsize \frac {1}{a^{4x}} \)

or

\(\frac{a^{4x}}{a^{4x}} \: – \: \frac {1}{a^{4x}} \\ = \frac {a^{4x} – 1}{a^{4x}} \)

SOLUTION (g)

:>\(\left (\frac {9a^2}{25}\right)^{^{- \frac {1}{2}}}\)

invert the fraction and change the negative index to positive

= \(\left (\frac {25}{9a^2}\right)^{^{ \frac {1}{2}}}\)

Note: 25 = 5 x 5 = \( \scriptsize 5^2\) and 3a x 3a \( \scriptsize = 3a^2= 9a^2\)

  \(\left (\frac {25}{9a^2}\right)^{ \frac {1}{2}} \\= \left(\frac {5^2}{\left(3a\right)^2}\right )^{ \frac {1}{2}}\)

= \( \left(\frac {5}{(3a)}\right)^{ 2\: \times \: \frac {1}{2}}\)

= \(  \frac{5}{3a}\)

SOLUTION (h)

\( \frac {-x^3}{(-x^6)( -x)}\scriptsize \: \times \: x^4 \)

 

multiply the numerator by xand open the brackets in the denominator

:> \( \frac {-x^3 \: \times \: x^4}{-x^6 \: \times \:  -x^1} \)

Using laws of logarithm \( \scriptsize x^a \: \times \: x^b = x^{a+b} \)

Alse Note: \( \scriptsize -x \: \times \: -x = +x \: or \: x \)

\( \frac {-x^{3+4}}{x^{6 + 1}} \\ = \frac {-x^7}{x^7}\)

Using laws of logarithm \(\frac {x^a}{x^b} \scriptsize = x^{a-b} \)

= \( \scriptsize -x^{7 – 7} \)

= \( \scriptsize -x^{0} \\ =  \scriptsize -1 \)

SOLUTION (i)

\( \left(\frac {8y^3}{(27p^6)}\right )^{^{- \frac {2}{3}}}\)

Invert the fraction and change the negative index to positive index.

i.e \( \left (\frac {27p^6}{8y^3}\right )^{^{\frac {2}{3}}}\)

= \(\left (\frac {3^3p^6}{2^3y^3}\right)^{^{\frac {2}{3}}}\)

= \( \frac {3^{^{\left(3 \: \times \:  \frac{2}{3}\right) }} \: \times \: p^{^{\left(6 \: \times  \: \frac{2}{3}\right )} }} {2^{^{\left(3 \: \times \:  \frac{2}{3}\right )}}\: \times \: y^{^{\left(3 \: \times \: \frac{2}{3}\right )}} }\)

= \( \frac{3^2 \: \times \: p^4}{2^2 \:\times \: y^2}\)

This can be simplified further as

= \(\left( \frac{3 p^2}{2y}\right)^{^2}\)

or

\(\left ( \frac{9 p^4}{4y^2}\right )\)

SOLUTION (j)

\( \scriptsize 32 ^{-0.4}\)

 

Change the Decimal to a Fraction

= \( \scriptsize 32 ^{^{ \normalsize -\frac{4}{10}}}\)

= \( \scriptsize \left(2^5\right ) ^{^{\normalsize  -\frac{2}{5}}}\)

= \( \scriptsize 2^{^{\normalsize \left(5\: \times \: -\frac{2}{5}\right)}} \)

= \( \scriptsize 2^{-2}\)

=\( \frac{1}{2^2} = \frac{1}{4}\)

SOLUTION (k)

\( \left(\frac{1}{8}\right) ^{^{ – 1 \frac{1}{3}}}\)

=\( (\frac{1}{8}) ^{^{- \frac{4}{3}}}\)

=\( (\frac{8}{1}) ^{^{ \frac{4}{3}}}\)

= \( \scriptsize(2^3)^{^{ \frac{4}{3}}} \\ \scriptsize = 2^{^{\normalsize 3 \: \times \: \frac{4}{3}}}\\ \scriptsize  = 2^4 \\ \scriptsize = 16\)

SOLUTION (l)

:> \( \left(\frac {9}{4} \scriptsize a^6\right )^{ ^{-\normalsize 1.5}}\)

1.5 = \( \frac{3}{2}\)

:> \( \left(\frac {9}{4} \scriptsize a^6\right)^{ ^{\normalsize -\frac{3}{2}}}\)

invert the fraction and change the negative index to positive index

= \( \left(\frac {4}{9a^6} \right ) ^{^{\normalsize \frac{3}{2}}}\)

= \( \left(\frac {2^2}{3^2 a^6} \right) ^{^{\normalsize\frac{3}{2}}}\)

= \( \frac {2^{^{\left(2 \: \times \: \frac{3}{2}\right)}}}{3^{^{\left(2\: \times \: \frac{3}{2}\right)}}\; a^{^{\left(6 \: \times \: \frac{3}{2}\right )}}} \)

= \( \left(\frac {2^3}{3^3 a^9} \right)\)

= \( \left (\frac {2}{3 a^3} \right)^{^3}\)

or \(\left (\frac {8}{9 a^9} \right)\)

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