Fundamental Laws of Logarithms – Part 2
3. For m, n, a > 0, a≠1; \({{\log }_{a}}\left( {{m}^{n}} \right)=n{{\log }_{a}}m\)
Proof: let \({{\log }_{a}}m=x\),
\({{a}^{x}}=m\ (or)\ {{\left( {{a}^{x}} \right)}^{n}}={{m}^{n}}\),
\({{a}^{xn}}={{m}^{n}}\),
\({{\log }_{a}}{{m}^{n}}=nx\),
\({{\log }_{a}}{{m}^{n}}=n{{\log }_{a}}m\),
Hence Proved.
4. For a > 1, if \({{\log }_{a}}{{x}_{2}}>{{x}_{1}}\ \ then\ \ 0<{{x}_{2}}<{{a}^{{{x}_{1}}}}\)
Proof: \({{\log }_{a}}{{x}_{2}}>{{x}_{1}}\),
\({{\log }_{a}}{{x}_{2}}>{{x}_{1}}{{\log }_{a}}a\),
\({{\log }_{a}}{{x}_{2}}>{{\log }_{a}}{{a}^{{{x}_{1}}}}\),
\({{x}_{2}}>{{a}^{{{x}_{1}}}}\) (since a > 1),
Hence proved.
5. For 0 < a < 1, if \({{\log }_{a}}{{x}_{2}}>{{x}_{1}}\ \ then\ \ 0<{{x}_{2}}<{{a}^{{{x}_{1}}}}\)
Proof: \({{\log }_{a}}{{x}_{2}}>{{x}_{1}}\),
\({{\log }_{a}}{{x}_{2}}>{{x}_{1}}{{\log }_{a}}a\),
\({{\log }_{a}}{{x}_{2}}>{{\log }_{a}}{{a}^{{{x}_{1}}}}\),
\(0<{{x}_{2}}<{{a}^{{{x}_{1}}}}\),
(since 0 < a < 1),
Hence proved.
6. If m, a, b > 0 and a ≠ 1, b ≠ 1, then \({{\log }_{a}}m=\frac{{{\log }_{b}}m}{{{\log }_{b}}a}\)
Proof: Let \({{\log }_{a}}m=x\). Then, \({{a}^{x}}=m\),
Taking log on both sides,
\({{\log }_{b}}{{a}^{x}}={{\log }_{b}}m\),
\(x{{\log }_{b}}a={{\log }_{b}}m\),
(since \(x={{\log }_{a}}m\,\)),
\({{\log }_{a}}m{{\log }_{b}}a={{\log }_{b}}m\),
\({{\log }_{a}}m=\frac{{{\log }_{b}}m}{{{\log }_{b}}a}\),
Put b = m,
\({{\log }_{a}}m=\frac{{{\log }_{m}}m}{{{\log }_{m}}a}\),
\({{\log }_{a}}m=\frac{1}{{{\log }_{m}}a}\).
Example: Find the value of \({{\log }_{2}}\left( 2\sqrt[3]{9}-2 \right)+{{\log }_{2}}\left( 12\sqrt[3]{3}+4+4\sqrt[3]{9} \right)\)
solution:
\({{\log }_{2}}\left( 2\sqrt[3]{9}-2 \right)+{{\log }_{2}}\left( 12\sqrt[3]{3}+4+4\sqrt[3]{9} \right)\),
\({{\log }_{2}}\left( {{\left( 72 \right)}^{1/3}}-{{\left( 8 \right)}^{1/3}} \right)+{{\log }_{2}}\left( {{\left( 72 \right)}^{2/3}}+{{\left( 8 \right)}^{2/3}}+{{\left( 72 \right)}^{1/3}}{{\left( 8 \right)}^{1/3}} \right)\),
\({{\log }_{2}}\left( \left( {{\left( 72 \right)}^{1/3}}-{{\left( 8 \right)}^{1/3}} \right)\left( {{\left( 72 \right)}^{2/3}}+{{\left( 8 \right)}^{2/3}}+{{\left( 72 \right)}^{1/3}}{{\left( 8 \right)}^{1/3}} \right) \right)\),
\({{\log }_{2}}\left( 72-8 \right)={{\log }_{2}}64\),
\({{\log }_{2}}{{2}^{6}}=6\).