Fundamental Laws of Logarithms- Part3

 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 \({{a}^{{{\log }_{b}}c}}=p\)\(\Rightarrow {{\log }_{b}}c={{\log }_{a}}p\),

\(\Rightarrow \frac{\log c}{\log b}=\frac{\log p}{\log a}\),

\(\Rightarrow \frac{\log a}{\log b}=\frac{\log p}{\log c}\),

\({{\log }_{b}}a={{\log }_{c}}p\),

\(p={{c}^{{{\log }_{b}}a}}\Rightarrow {{a}^{{{\log }_{b}}c}}={{c}^{{{\log }_{b}}a}}\).