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Please help me proving the basic logarithmic identity $\log_3 12=1+\log_5 4\cdot \log_6 5\cdot \log_3 6$

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3 Answers

up vote 4 down vote accepted

Before we prove the identity mention some logaritmic identity:

1) $\log_a b=\frac{\log_x b}{\log_x a}$

2) $\log_a xy=\log_a x + \log_a y$

3) $\log_a a=1$

$\log_3 12=1+\log_5 4\cdot \log_6 5\cdot \log_3 6$

$\log_3 12=1+\frac{\log 4}{\log 5}\cdot \frac{\log 5}{\log 6}\cdot \frac{\log 6}{\log 3}$

$\log_3 12=1+\frac{\log 4}{\log 3}$

$\log_3 12=\log_3 3+\log_3 4$

$\log_3 12=\log_3 12$

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Using the power rule: $n\log_a(x)=\log_a(x^n)$

and the product rule $\log_a(a)+\log_a(b)=\log_a(ab)$ you get:

$$1+\log_5 4\cdot \log_6 5\cdot \log_3 6\\= 1+ \log_6 (5^{\log_5 4})\cdot \log_3 6$$

$$=1+ \log_6 (4)\cdot \log_3 6\\=1+ \cdot \log_3( 6^{\log_6 (4)})$$

$$=1+\log_3( 4)\\=\log_3(3)+\log_3( 4)\\=\log_3(12)$$

This avoids a bit of log writing using the other proposed identity, but not that much. In the end it's a matter of taste.

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$1+\log_54\cdot \log_65\cdot \log_36$ $=1+\frac{\log4\cdot \log 5\cdot \log6}{\log 5\cdot \log 6 \cdot \log 3}$ $=1+\frac{\log4}{\log3}=\frac{\log3}{\log3}+\frac{\log4}{\log3}$ $=\log_33+\log_34=\log_3(3.4)=\log_312$

as $$\log_xy=\frac{\log_zy}{\log_zx}$$ and $$\log_zx+\log_zy=\log_zxy$$ where $z$ is positive real number $\neq 1$ as $\log_1x$ is not defined.

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