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SS2: MATHEMATICS - 1ST TERM

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  1. Logarithms of Numbers Less Than One
    2 Topics
  2. Theory of Logarithms
    6 Topics
    |
    1 Quiz
  3. Approximation & Accuracy
    5 Topics
    |
    1 Quiz
  4. Sequence
    2 Topics
  5. Series
    2 Topics
    |
    1 Quiz
  6. Geometric Progression
    2 Topics
    |
    1 Quiz
  7. Quadratic Equations II
    2 Topics
  8. Quadratic Equations III
    1 Topic
  9. Quadratic Equations IV
    3 Topics
    |
    1 Quiz

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SS2: MATHEMATICS – 1ST TERM Theory of Logarithms Changing the Base of a logarithm
Lesson 2, Topic 3
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Changing the Base of a logarithm

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Topic Content:

  • Changing the base of a logarithm
  • Worked Examples – Logarithms

Note that it is possible to change the base of a given logarithm to a more convenient base.

Suppose logqP = y, then  \( \scriptsize q^y = P \)

Taking logs to base a of both sides of 

qy = p

\( \scriptsize log_a q^y = log_a P \)

\( \scriptsize y log_a q = log_a P \)

i.e y = \( \frac{log_aP}{log_aq} \)

∴lolq P = \( \frac{log_aP}{lolg_aq} \) (where p, q are positive real \( \scriptsize \neq \) numbers  1).

If  a = P  then,

logq a = \( \frac{log_aa}{log_aq} \)

i.e logq a = \( \frac{1}{log_aq} \)

Worked Examples 2.3.1:

Evaluate the following: (Try to work these examples out on your own using the laws of logarithms and then check the solutions by clicking ‘view solution’ )

(a) \( \scriptsize 3 \log 4 + \log 2\)

View Solution

(b) \( \scriptsize 3 \log 2 + \log 20 \; – \log1.6\)

View Solution

(c) \( \scriptsize 2 \: – \: 2 \log 5\)

View Solution

(d) \( \frac {\log 8 \: – \: \log 4}{\log 6 \: – \: \log 3} \)

View Solution

(e) \( \frac{1}{2}\scriptsize \log \normalsize \frac{25}{4}\scriptsize \: – \: 2 \log \normalsize \frac{4}{5} \scriptsize \: + \: \log \normalsize \frac{320}{125} \)

View Solution

(f) \( \scriptsize \log \normalsize \frac{30}{16}\scriptsize \: – \: 2 \log \normalsize \frac{5}{9} \scriptsize\: + \: \log \normalsize \frac{400}{243} \)

View Solution

(g) \( \scriptsize \log_{36} 6\: +\: \log_3 27\: -\: \log_9 3 \)

View Solution

(h) \( \scriptsize \log_{3} 64\: \times \: \log_8 243 \)

View Solution

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SOLUTION (a)

Question

⇒ \( \scriptsize 3 \log 4 + \log 2\)

Solution

Using the laws of logarithm

:> \(\scriptsize \log_{a} \left ( M^{n} \right ) = n \log_{a} M \)

⇒\( \scriptsize 3 \log 4 = \log 4^3 \)

⇒ \( \scriptsize \log 4^3 \: +\:  \log 2\)

\(\scriptsize \log_a M + \log_aN = \log_a MN  \)

⇒ \( \scriptsize\log 4^3 \: +\:  \log 2 \\ \scriptsize =  \log 4^{3} \: \times \:  2 \)

=  \( \scriptsize \log 64 \: \times \: 2 \)

= \( \scriptsize \log 128 \)

SOLUTION (b)

Question

⇒ \( \scriptsize 3 \log 2 \: + \: \log 20 \:  – \:  \log1.6\)

Solution

⇒ \( \scriptsize  \log 2^3 \: + \:  \log 20  \: – \: \log \normalsize \frac {16}{10}\)

Laws of Logarithm

⇒\( \scriptsize \log_a MN = \log_a M \: +\: \log_aN \\  \)

⇒ \( \scriptsize \log_a \normalsize\left (\frac{M}{N} \right ) \scriptsize = \log_a M \: – \: \log_aN \)

⇒ \( \scriptsize  \log \normalsize \frac {2^3 \: \times \: 20}{ \frac {16}{10}}\)

⇒ \( \scriptsize  \log \normalsize \frac {8 \: \times \: 20}{ \frac {16}{10}}\)

⇒ \( \scriptsize  \log  \left [ \scriptsize  8 \: \times \:  20 \:\times \: \normalsize \frac {10}{16} \right ] \)

⇒ \( \scriptsize  \log \left ( \scriptsize 10 \: \times \: 10 \right)  \)

⇒ \( \scriptsize  \log 10^2 \\ \scriptsize =  2 \log_{10} 10 \)

From Laws of logarithm 

⇒ \(\scriptsize \log_{a}a = 1 \\ \scriptsize\therefore \log_{10} 10 = 1 \)

= 2 x 1 = 2

SOLUTION (c)

Question

⇒ \( \scriptsize 2 \: –  \: 2 \log 5\)

Solution

⇒ \( \scriptsize 2 \: \times \: 1 – \: 2 \log 5\)

⇒ \( \scriptsize 2 \log 10 \: –  \: \log 5^2\)

⇒ \( \scriptsize  \log 100 \: – \: \log 25\)

⇒ \( \scriptsize \log \normalsize \left ( \frac {100}{25} \right)\)

⇒ \( \scriptsize  \log 4\)

SOLUTION (d)

Question

⇒ \( \frac {\log 8 \; – \;\log 4}{\log 6 \; – \; \log 3} \)

Solution

⇒ \( \frac {\log \frac{8}{4}}{\log \frac{6}{3}} \)

⇒ \( \frac {\log 2}{\log 2}  \scriptsize =  1 \)

SOLUTION (e)

Question

⇒ \( \frac{1}{2}\scriptsize \log \normalsize \frac{25}{4}\scriptsize \: – \:2  \log \normalsize \frac{4}{5} \scriptsize \:+\: \log \normalsize \frac{320}{125} \)

Solution

⇒ \(\scriptsize \log \normalsize \left ( \frac{25}{4}\right)^{\frac{1}{2}} \scriptsize \: – \: \log \normalsize \left ( \frac{4}{5}\right)^{2} \scriptsize\: + \: \log \normalsize \frac{320}{125} \)

⇒ \(\scriptsize \log \normalsize \frac{\sqrt{25}}{\sqrt{4}} \scriptsize \: – \: \log \normalsize \left ( \frac{4}{5}\right)^{2} \scriptsize\: + \: \log \normalsize \frac{320}{125} \)

⇒ \(\scriptsize \log \normalsize \frac{5}{2} \scriptsize\: -\: log \normalsize \frac{16}{25} \scriptsize \:+ \:\log \normalsize \frac{320}{125} \)

⇒ \(\scriptsize \log  \left [\normalsize \frac{5}{2}\: \div \:  \normalsize \frac{16}{25} \: \times \: \normalsize \frac{320}{125} \right] \)

⇒ \(\scriptsize \log  \left [\normalsize \frac{5}{2} \: \times \:  \normalsize \frac{25}{16}\:  \times \:   \normalsize \frac{320}{125} \right] \)

⇒ \(\scriptsize \log  \left [ \normalsize \frac{25}{16}\:  \times \:   \normalsize \frac{160}{25} \right] \)

= \(\scriptsize \log_{10}  10  \\ \scriptsize = 1\)

SOLUTION (f)

Question

⇒ \( \scriptsize \log \normalsize \frac{30}{16}\scriptsize \: – \: 2 \log \normalsize \frac{5}{9} \scriptsize \:+ \: \log \normalsize \frac{400}{243} \)

Solution

⇒ \( \scriptsize \log \normalsize \frac{30}{16}\scriptsize \: – \: \log \normalsize \left ( \frac{5}{9} \right)^2 \scriptsize \: +\: \log \normalsize \frac{400}{243} \)

⇒ \( \scriptsize \log \normalsize \frac{30}{16}\scriptsize \: – \: \log \normalsize \frac{25}{81} \scriptsize \:+ \: \log \normalsize \frac{400}{243} \)

⇒ \( \scriptsize \log \left [\normalsize \frac{30}{16} \scriptsize \: \div \: \normalsize  \frac{25}{81} \scriptsize  \: \times \: \normalsize  \frac{400}{243}\right] \)

⇒ \( \scriptsize \log \left [\normalsize \frac{30}{16} \scriptsize \: \times \:  \normalsize\frac{81}{25} \scriptsize  \: \times \:  \normalsize  \frac{400}{243}\right] \)

= \( \scriptsize \log_{10} 10\\ \scriptsize  = 1 \)

SOLUTION (g)

Question

⇒ \( \scriptsize \log_{36} 6\; +\; log_3 27\; -\; log_9 3 \)

Solution

Change the log to base 3

Change of base rule \( \scriptsize \log_{b} (a) = \normalsize \frac{\log_x (a)}{\log_x (b)} \)

⇒ \( \scriptsize \log_{36} (6) = \normalsize \frac{\log_3 (6)}{\log_3(36)} \)

⇒ \( \scriptsize \log_{9} (3) = \normalsize \frac{\log_3 (3)}{\log_3 (9)} \)

Substitute these values into the original equation

⇒ \( \frac{\log_3 (6)}{\log_3(36)} \scriptsize \: +\: \log_3 27\: -\: \normalsize \frac{\log_3 (3)}{\log_3 (9)} \)

⇒ \( \frac{\log_3 (6)}{\log_3(6^2)} \scriptsize \: +\: \log_3 27\: -\: \normalsize \frac{\log_3 (3)}{\log_3 (3^2)} \)

⇒ \( \frac{\log_3 (6)}{\log_3(6^2)} \scriptsize \: +\: \log_3 3^3\: -\: \normalsize \frac{\log_3 (3)}{\log_3 (3^2)} \)

⇒ \( \frac{\log_3 (6)}{\log_3(6^2)} \scriptsize \: +\: 3\log_3 3\: -\: \normalsize \frac{\log_3 (3)}{\log_3 (3^2)} \)

⇒ \( \frac{\log_3 (6)}{2\log_3(6)} \scriptsize \: +\: 3 \: \times \: 1\: -\: \normalsize \frac{\log_3 (3)}{2\log_3 (3)} \)

Note: \(\normalsize  \frac{\log_3 (6)}{\log_3(6)} \scriptsize = 1 \: and \: \normalsize \frac{\log_3 (3)}{\log_3 (3)} \scriptsize = 1  \)

⇒ \( \normalsize \frac {1}{2} \scriptsize  \: + \: 3 \:  –  \: \normalsize \frac{1}{2}\)

⇒ \( \scriptsize 3 \normalsize \frac {1}{2}  \:  –  \: \normalsize \frac{1}{2}\)

= 3

SOLUTION (h)

Question

⇒ \( \scriptsize \log_{3} 64\: \times \: \log_8 243 \)

Solution

Change the log to base 3

Change of base rule: \( \scriptsize \log_{b} (a) = \normalsize \frac{\log_x (a)}{\log_x (b)} \)

⇒ \( \scriptsize \log_{8} (243) = \normalsize \frac{\log_3 (243)}{\log_3(8)} \)

Substitute these values into the original equation

⇒ \( \scriptsize \log_{3} 64\: \times \: \normalsize \frac{\log_3 (243)}{\log_3(8)} \)

⇒ \( \scriptsize \log_{3} 8^2\: \times \: \normalsize \frac{\log_3 (3^5)}{\log_3(8)} \)

⇒ \( \scriptsize 2\log_{3} 8\: \times \: \normalsize \frac{5\log_3 (3)}{\log_3(8)} \)

The logs can cancel out and \( \scriptsize \log_3 3 = 1 \)

= 2 x 5 x 1 = 10

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