Fundamental Laws of Logarithms- Part3
7. For a, n > 0 and a ≠ 1; \({{a}^{{{\log }_{a}}n}}=n\)
Proof:
Let \({{\log }_{a}}n=x\). Then \({{a}^{x}}=n\).
Therefore, \({{a}^{{{\log }_{a}}n}}=n\),
Examples:
(1) \({{3}^{{{\log }_{3}}8}}=8\)
(2) \({{2}^{3{{\log }_{2}}6}}={{2}^{{{\log }_{2}}{{6}^{3}}}}={{6}^{3}}\)
(3) \({{5}^{-3{{\log }_{5}}2}}={{5}^{{{\log }_{5}}{{2}^{-3}}}}=\frac{1}{{{2}^{3}}}\)
8. \({{\log }_{{{a}^{q}}}}{{n}^{p}}=\frac{p}{q}{{\log }_{a}}n=y\), where a, n > 0, a ≠ 1
Proof:
Let \({{\log }_{{{a}^{q}}}}{{n}^{p}}=x\ \ and\ \ {{\log }_{a}}n=y.\)then,
\({{\left( {{a}^{q}} \right)}^{x}}={{n}^{p}}\ and\ {{a}^{y}}=n\),
\(\left( {{a}^{qx}} \right)={{n}^{p}}\ and\ {{a}^{y}}=n\),
\(\left( {{a}^{qx}} \right)={{n}^{p}}\ and\ {{\left( {{a}^{y}} \right)}^{p}}={{n}^{p}}\),
\({{a}^{qx}}={{a}^{yp}}\),
qx = yp,
⇒ x = (p/q)y,
⇒ \({{\log }_{{{a}^{q}}}}{{n}^{p}}=\frac{p}{q}{{\log }_{a}}n\).
Example: What is greater \(x={{\log }_{3}}5\) (or) \(y={{\log }_{17}}25\)
Solution:
\(y={{\log }_{17}}25\),
\(\frac{1}{y}={{\log }_{25}}17\),
\(={{\log }_{{{5}^{2}}}}17\),
\(=\frac{1}{2}{{\log }_{5}}17\)….(1),
\(x={{\log }_{3}}5\),
\(\frac{1}{x}={{\log }_{5}}3\)….(2),
From equation (1) and (2),
\(\frac{1}{y}>\frac{1}{x}\) ,
x > y.
9. \({{a}^{{{\log }_{b}}c}}={{c}^{{{\log }_{b}}a}}\)
Proof: Let \({{a}^{{{\log }_{b}}c}}=p\)\(\Rightarrow {{\log }_{b}}c={{\log }_{a}}p\),
\(\Rightarrow \frac{\log c}{\log b}=\frac{\log p}{\log a}\),
\(\Rightarrow \frac{\log a}{\log b}=\frac{\log p}{\log c}\),
\({{\log }_{b}}a={{\log }_{c}}p\),
\(p={{c}^{{{\log }_{b}}a}}\Rightarrow {{a}^{{{\log }_{b}}c}}={{c}^{{{\log }_{b}}a}}\).