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Laws of logarithms

Laws of Logarithms

The following are the key laws of logarithms
  • $\log_{a}(0) = 1$
  • $\log_{a}(a) = 1$
  • $\log_{a}(bc) = \log_{a}(b) + \log_a(c)$
  • $\log_{a}\left(b^c\right) = c\log_{a}(b)$
  • $\log_{a}\left(\frac{1}{b}\right) = -\log_{a}(b)$
  • $\log_{a}\left(\frac{b}{c}\right) = \log_{a}(b) - \log_a(c)$
  • $\log_{a}\left(\sqrt[c]{b}\right) = \frac{1}{c}\log_{a}(b)$
  • $\log_{a}(b) = \frac{\log_{c}(b)}{\log_{c}(a)}$

It is worth a moment to cross reference them with the laws of indices! 

The final law of logs is the follow equivalence:

$$\log_a(b) = c \Leftrightarrow a^c = b$$

This law allows us to insert or remove logarithms from an equation. 

Logarithms 1

Using logarithms, solve the following equation $$ 3\times 2^{2x+1} = 15 $$
solution - press button to display
$$ \begin{align} 3\times 2^{2x+1} &= 15\\ 2^{2x+1} &= 5 \\ \log_2{2^{2x+1}} &= \log_2{5} \\ 2x+1 &= \log_2{5} \\ x &= \frac{\log_2{5}-1}{2} \end{align} $$

Logarithms 2

Solve the equation $$ 2^{3+x} = 5^{2x-4} $$ expressing your answer in terms of $\log_2(5)$

solution - press button to display

$$ \begin{align} 2^{3+x} &= 5^{2x-4} \\ \log_2\left(2^{3+x}\right) &= \log_2\left(5^{2x-4}\right) \\ 3+x &= (2x-4)\log_2(5) \\ 3+x &= 2\log_2(5)x - 4\log_2(5) \\ 3 + 4\log_2(5) &= (2\log_2(5) - 1)x \\ x &= \frac{3+4\log_2(5)}{-1 + 2\log_2(5)} \end{align}$$