2.1 Proofs of all the above properties

1.1 Case 1: When $a=1$
1.2 Case 2: When $a=0$
1.3 Case 3: When $N<0$

3.0 Relation between common logarithm $\left( {{{\log }_{10}}x} \right)$ and Natural logarithm $\left( {{{\log }_e}x} \right)$
2.1 Proofs of all the above properties
2.2 System of logarithms

From the definition of logarithmic function$$\begin{equation} \begin{aligned} {\log _a}N = x \\ {a^x} = N \\\end{aligned} \end{equation} $$ Therefore for, $$\begin{equation} \begin{aligned} N = 1 \\ x = 0 \\\end{aligned} \end{equation} $$ Hence Proved.

From the definition of logarithmic function$$\begin{equation} \begin{aligned} {\log _a}N = x \\ {a^x} = N \\\end{aligned} \end{equation} $$ Therefore for, $$\begin{equation} \begin{aligned} N = a \\ x = 1 \\\end{aligned} \end{equation} $$ Hence Proved.

Let, $$\begin{equation} \begin{aligned} {\log _a}x = A\quad or\quad {a^A} = x\quad ...(i) \\ {\log _a}y = B\quad or\quad {a^B} = y\quad ...(ii) \\ {\log _a}xy = C\quad or\quad {a^C} = xy\quad ...(iii) \\\end{aligned} \end{equation} $$ So, we can write the above property as,$$C = A + B\quad ...(iv)$$

Multiplying equation $(i)$ & $(ii)$ we get,$$\begin{equation} \begin{aligned} {a^C} = {a^{A + B}} \\ C = A + B \\\end{aligned} \end{equation} $$ Hence Proved.

Let, $$\begin{equation} \begin{aligned} {\log _a}x = A\quad or\quad {a^A} = x\quad ...(i) \\ {\log _a}y = B\quad or\quad {a^B} = y\quad ...(ii) \\ {\log _a}\left( {\frac{x}{y}} \right) = C\quad or\quad {a^C} = \left( {\frac{x}{y}} \right)\quad ...(iii) \\\end{aligned} \end{equation} $$ So, we can write the above property as,$$C = A - B\quad ...(iv)$$

Dividing equation $(i)$ by $(ii)$ we get,$$\begin{equation} \begin{aligned} {a^C} = {a^{A - B}} \\ C = A- B \\\end{aligned} \end{equation} $$ Hence Proved.

Let, $$\begin{equation} \begin{aligned} {\log _a}x = A\quad or\quad {a^A} = x\quad ...(i) \\ {\log _a}{x^n} = B\quad or\quad {a^B} = {x^n}\quad ...(ii) \\\end{aligned} \end{equation} $$ From equation $(i)$ & $(ii)$ we get, $$\begin{equation} \begin{aligned} {a^B} = {\left( {{a^A}} \right)^n} \\ {a^B} = {a^{An}} \\ B = An \\\end{aligned} \end{equation} $$ or $${\log _a}{x^n} = n{\log _a}x$$ Hence Proved.

As $2k$ is always even. So, $${x^{2k}} > 0$$ Therefore, $x$ can be either positive or negative but cannot be zero. So, that the logarithm is defined.

Mathematically, $$x:R - \left\{ 0 \right\}$$ So, by using property 5. $${\log _a}{x^{2k}} = 2k{\log _a}\left| x \right|$$ Hence Proved.

Let, $$\begin{equation} \begin{aligned} {\log _a}x = A\quad ...(i) \\ {a^A} = x\quad ...(ii) \\ {\log _{{a^m}}}x = B\quad ...(iii) \\ {a^{mB}} = x\quad ...(iv) \\\end{aligned} \end{equation} $$ From equation $(iii)$ & $(iv)$ we get, $$\begin{equation} \begin{aligned} {a^{mB}} = {a^A} \\ mB = A \\ m{\log _{{a^m}}}x = {\log _a}x \\ {\log _{{a^m}}}x = \frac{1}{m}{\log _a}x \\\end{aligned} \end{equation} $$ Hence Proved.

As $2k$ is always even. So, $${a^{2k}} > 0$$ Therefore, $x$ can be either positive or negative but cannot be zero. So, that the logarithm is defined.

Mathematically, $$x:R - \left\{ 0 \right\}$$ So, by using property 5. $${\log _a}{x^{2k}} = 2k{\log _a}\left| x \right|$$ Hence Proved.

Let, $$\begin{equation} \begin{aligned} {\log _a}b = A\quad ...(i) \\ {a^A} = b\quad ...(ii) \\ {\log _{{a^n}}}{b^m} = B\quad ...(iii) \\ {a^{nB}} = {b^m}\quad ...(iv) \\\end{aligned} \end{equation} $$ From equation $(iii)$ & $(iv)$ we get, $$\begin{equation} \begin{aligned} {a^{nB}} = {a^{Am}} \\ nb = Am \\ n{\log _{{a^n}}}{b^m} = m{\log _a}b \\ {\log _{{a^n}}}{b^m} = \frac{m}{n}{\log _a}b \\\end{aligned} \end{equation} $$ Hence Proved.

Let, $$\begin{equation} \begin{aligned} {\log _a}b = A\quad ...(i) \\ {a^A} = b\quad ...(ii) \\ \\ {\log _c}b = B\quad ...(iii) \\ {c^B} = b\quad ...(iv) \\ \\ {\log _c}a = C\quad ...(v) \\ {c^C} = a\quad ...(vi) \\\end{aligned} \end{equation} $$ From equation $(ii)$ & $(vi)$ we get,$$b = {a^{AC}}\quad ...(vii)$$ From equation $(iv)$ & $(vii)$ we get, $$\begin{equation} \begin{aligned} {a^B} = {a^{AC}} \\ B = AC \\ A = \frac{B}{C} \\\end{aligned} \end{equation} $$ or $${\log _a}b = \frac{{{{\log }_c}b}}{{{{\log }_c}a}}$$ Hence Proved.

Let, $$\begin{equation} \begin{aligned} {\log _a}b = A\quad ...(i) \\ {a^A} = b\quad ...\left( {ii} \right) \\ {\log _b}a = B\quad ...(iii) \\ {b^B} = a\quad ...\left( {iv} \right) \\\end{aligned} \end{equation} $$ From equation $(ii)$ & $(iv)$ we get, $$\begin{equation} \begin{aligned} {a^{AB}} = {a^1} \\ AB = 1 \\ A = \frac{1}{B} \\ {\log _a}b = \frac{1}{{{{\log }_b}a}} \\\end{aligned} \end{equation} $$ Hence Proved.

Let, $$\begin{equation} \begin{aligned} {\log _a}b = A\quad ...(i) \\ {a^A} = b\quad ...\left( {ii} \right) \\ {\log _b}C = B\quad ...(iii) \\ {b^B} = C\quad ...\left( {iv} \right) \\ {\log _c}a = C\quad ...(v) \\ {c^C} = a\quad ...\left( {vi} \right) \\\end{aligned} \end{equation} $$ From equation $(ii)$, $(iv)$ & $(vi)$ we get, $$\begin{equation} \begin{aligned} {c^1} = {c^{ABC}} \\ ABC = 1 \\ {\log _a}b \times {\log _b}c \times {\log _c}a = 1 \\\end{aligned} \end{equation} $$ Hence Proved.

Let, $$\begin{equation} \begin{aligned} {\log _a}x = A \\ {a^A} = x \\\end{aligned} \end{equation} $$ Also, ${a^{{{\log }_a}x}} = x$ can be written as ${a^A} = x$