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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
Lesson Progress
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The various laws of indices are also of importance in solving simple exponentials given below:

Example

Solve the following equations:

(a) \( \scriptsize 5^{2x – 4} = 25 ^{-x – 8}\)

(b) \( \scriptsize 8^x = 0.125 \)

(c) \( \scriptsize p^x = \normalsize \frac{ \sqrt[4] {p^5}\: \times \: p^{- \frac{1}{4}}}{ \left(\sqrt [3] {p}\right)^2} \)

(d) \( \frac {1}{4} \scriptsize \; of \; 64x = 16^{3x}\)

(e) \( \scriptsize 9^{x – 3} = 27 ^{x – 5}\)

(f) \( \scriptsize 9^{2x + 1} = \normalsize \frac {81 ^{x – 3}}{3^x}\)

(g) \( \scriptsize 7(8^{x + 1}) = 448\)

(h) \( \left(\frac {5}{6}\right)^{ \frac{1}{2}} = \left(\frac {6}{5}\right)^{x – 1}\)

(i) \( \frac {64}{27} = \left(\frac {3}{4}\right )^{^{\normalsize x – 1}}\)

Evaluation:

Solve the following;

(i) \( \scriptsize 9^{2x} = \normalsize \frac{81^{x-2}}{3^x} \)

(ii)\( \scriptsize 2^{2x} \: – \: 3\: \times \: 2^x \: + \: 2 = 0 \)

SOLUTION (a)

:> \( \scriptsize 5^{2x – 4} = 25 ^{-x – 8}\)

We know \( \scriptsize 25 = 5^2 \)

\( \scriptsize 5^{2x – 4} = 5^{2(-x – 8)}\)

:> \( \scriptsize 5^{2x – 4} = 5^{-2x\: – \:16}\)

Since we have equal base, we can equate the like terms

i.e. \( \scriptsize 2x \: – \: 4 = \: -2x \: – \: 16 \)

Collecting like terms

i.e \( \scriptsize 2x \: + \: 2x = \: -16 \: +\: 4 \)

i.e. \( \scriptsize 4x  = \: -12 \)

Divide both sides by 4

:> \( \frac{4x}{4} = \frac{-12}{4} \)

i.e. \(\scriptsize x = \: -3 \)

SOLUTION (b)

\( \scriptsize 8^x = 0.125 \)

Change the decimal to fraction

\( \scriptsize 8^x = \normalsize \frac {125}{1000} \)

i.e  \( \scriptsize 8^x = \normalsize \frac {1}{8} \)

\( \scriptsize 8^x = 8^{-1} \)

Since we have equal base we can equate the powers

x = -1

SOLUTION (c)

:> \( \scriptsize p^x = \normalsize \frac{ \sqrt[4] {p^5}\: \times \: p^{- \frac{1}{4}}}{ \left(\sqrt [3] {p}\right)^2} \)

\( \scriptsize p^x = \normalsize \frac{ p^{ 5 \: \times \: \frac {1}{4}} \: \times \: p^{ -\frac{1}{4}}}{ \left(\sqrt [3] {p}\right)^2} \)

\( \scriptsize p^x = \normalsize \frac{ p^{ \frac {5}{4}} \: \times \: p^{- \frac{1}{4}}}{ \left(\sqrt [3] {p}\right)^2} \)

:> \( \scriptsize p^x = \normalsize \frac{ p^{ \frac {5}{4}} \: \times \: p^{- \frac{1}{4}}}{ \left(p^{ \frac{1}{3}}\right) ^{^2}} \)

:> \( \scriptsize p^x = \normalsize \frac{ p^{ \frac {5}{4}} \: \times \: p^{ -\frac{1}{4}}}{ p^{ \frac{1}{3} \: \times \: 2}} \)

\( \scriptsize p^x = \normalsize \frac{ p^{^{ \normalsize \frac {5}{4}\:  – \:\frac{1}{4}}}}{ p^{^{ \normalsize\frac{2}{3}}}} \)

= \( \scriptsize p^x = \normalsize \frac{ p^{^{ \normalsize \frac {4}{4}}}}{ p^{^{ \normalsize\frac{2}{3}}}} \)

\( \scriptsize p^x = \normalsize \frac{ p^{ 1}}{ p^{ \frac{2}{3}}} \)

\( \scriptsize p^x = p^{ \left( 1 \:-\: \normalsize\frac{2}{3}\right)} \)

:>  \( \scriptsize p^x =  p^{^{\left(\normalsize \frac{3}{3} \:-\: \frac{2}{3}\right) }} \)

\( \scriptsize p^x =  p^{\normalsize\frac{1}{3}} \)

(equating powers)

Therefore, \( \scriptsize x = \normalsize \frac {1}{3}\)

SOLUTION (d)

:> \( \frac {1}{4} \scriptsize \: of \: 64x = 16^{3x}\)

:> \( \scriptsize 4^{-1} \times (4^3)^x = (4^2)^{3x}\)

(equating powers)

:> \( \scriptsize 4^{3x – 1} = 4^{6x}\)

Therefore, \( \scriptsize 3x \: – \: 1 = 6x \)

Collect like terms

:> \( \scriptsize 3x \: -\: 6x = 1\)

i.e. \( \scriptsize -3x = 1 \)

\( \scriptsize x = \: – \normalsize \frac {1}{3} \)

SOLUTION (e)

\( \scriptsize 9^{x – 3} = 27 ^{x – 5}\)

\( \scriptsize 3^{2(x – 3)} = 3^{3(x – 5)}\)

equating powers

i.e.   2(x – 3) = 3(x – 5)

i.e. 2x – 6 = 3x – 15

(collect like terms)

2x 3x = 15 + 6

–x = 9

divide both sides by -1

x = 9

SOLUTION (f)

:> \( \scriptsize 9^{2x + 1} = \normalsize \frac {81 ^{x – 3}}{3^x}\)

:> \( \scriptsize 3^{2 (2x + 1)} = \normalsize \frac {3 ^{4(x – 3)}}{3^x}\)

:> \( \scriptsize 3^{2 (2x + 1)} = 3^{4(x – 3) – x}\)

(equating powers)

2(2x + 1) = 4(x – 3) – x

open brackets

4x + 2 = 4x – 12 – x

4x + 2 = 4x – x – 12

i.e. 4x + 2 = 3x – 12

(collect like terms)

4x 3x = 12 2

∴ x = –14

SOLUTION (g)

:> \( \scriptsize 7(8^{x + 1}) = 448\)

Divide both sides by 7

:> \( \frac {7(8^{x + 1})}{7} = \frac {448}{7}\)

:> \( \scriptsize 8^{x + 1} = 64 \)

:> \( \scriptsize 8^{x + 1} = 8^2 \)

(equating powers)

i.e.     x + 1 = 2

x = 2 – 1

x = 1

SOLUTION (h)

:> \( \left(\frac {5}{6}\right)^{ \frac{1}{2}} = \left(\frac {6}{5}\right)^{x – 1}\)

Recall 

:> \(\left( \frac {5}{6}\right)^{ \frac{1}{2}} = \left(\frac {6}{5}\right)^{ – \frac{1}{2}}\)

\(\left(\frac {6}{5}\right)^{ – \frac{1}{2}} = \left(\frac {6}{5}\right)^{x – 1}\)

Therefore, equating powers, we have

:> \( – \frac{1}{2}  \scriptsize = x\:  – \:  1\)

:> \( \scriptsize x = 1\:  –  \: \normalsize \frac{1}{2}\)

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

SOLUTION (i)

:> \( \frac {64}{27} = \left(\frac {3}{4}\right )^{^{\normalsize x – 1}}\)

Invert the fraction and change the index to negative index.

i.e \( \frac {4^3}{3^3} = \left(\frac {4}{3}\right )^{-(x – 1)}\)

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

(equating powers)

i.e.  3 = -(x – 1)

3 = -x + 1

x = -2

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