Fundamental Laws of Logarithms – Part 1
1.For m, n > 0, a ≠ 1; \({{\log }_{a}}\left( mn \right)={{\log }_{a}}m+{{\log }_{a}}n\)
Proof: Let \({{\log }_{a}}m=x\) and \({{\log }_{a}}n=y\),
\({{\log }_{a}}m=x\Rightarrow m={{a}^{x}}\)…(1),
\({{\log }_{a}}n=y\Rightarrow n={{a}^{y}}\)….(2),
From equation (1) and (2),
\(mn={{a}^{x}}{{a}^{y}}\),
\(mn={{a}^{x+y}}\),
\({{\log }_{a}}\left( mn \right)=x+y\),
\({{\log }_{a}}\left( mn \right)={{\log }_{a}}m+{{\log }_{a}}n\),
Hence proved.
In general, if \({{x}_{1}},{{x}_{2}},…,{{x}_{n}}\) are positive real numbers, then,
\({{\log }_{a}}\left( {{x}_{1}}{{x}_{2}}…{{x}_{n}} \right)={{\log }_{a}}{{x}_{1}}+{{\log }_{a}}{{x}_{2}}+….+{{\log }_{a}}{{x}_{n}}\),
2. For m, n, a>0, a ≠ 1; \({{\log }_{a}}\left( \frac{m}{n} \right)={{\log }_{a}}m-{{\log }_{a}}n\).
Proof: Let \({{\log }_{a}}m=x\) and \({{\log }_{a}}n=y\)
\({{\log }_{a}}m=x\Rightarrow m={{a}^{x}}\)…(1),
\({{\log }_{a}}n=y\Rightarrow n={{a}^{y}}\)….(2),
From equation (1) and (2),
\(\frac{m}{n}=\frac{{{a}^{x}}}{{{a}^{y}}}\),
\(\frac{m}{n}={{a}^{x-y}}\),
\({{\log }_{a}}\left( \frac{m}{n} \right)=x-y={{\log }_{a}}m-{{\log }_{a}}n\),
\({{\log }_{a}}\left( \frac{m}{n} \right)={{\log }_{a}}m-{{\log }_{a}}n\),
Hence proved.
Example: Solve \({{\log }_{e}}\left( \frac{a+b}{2} \right)=\left( {{\log }_{e}}a+{{\log }_{e}}b \right)\)
Solution:
\({{\log }_{e}}\left( \frac{a+b}{2} \right)=\left( {{\log }_{e}}a+{{\log }_{e}}b \right)\),
\({{\log }_{e}}\left( \frac{a+b}{2} \right)={{\log }_{e}}\left( ab \right)\),
\(\left( \frac{a+b}{2} \right)=\left( ab \right)\),
\(a+b=2ab\).