The logarithm of the number N\mathrm{N} to the base ‘ a\mathrm{a} ‘ is the exponent indicating the power to which the base ‘a’ must be raised to obtain the number NN. This number is designated as log_(a)N\log _{a} N.
(a) log_(a)N=x\log _{a} \mathrm{~N}=\mathrm{x}, read as log\log of N\mathrm{N} to the base a<=>a^(x)=N\mathrm{a} \Leftrightarrow \mathrm{a}^{\mathrm{x}}=\mathrm{N} If a=10\mathrm{a}=10 then we write log N\log \mathrm{N} or log_(10)N\log _{10} \mathrm{~N} and if a=e\mathrm{a}=e we write ln N\ln \mathrm{N} or log_(e)N\log _{e} \mathrm{~N} (Natural log\log )
(b) Necessary conditions: N > 0;a > 0;a!=1\mathrm{N}>0 ; \mathrm{a}>0 ; \mathrm{a} \neq 1
(c) log_(a)1=0\log _{a} 1=0
(d) log_(a)a=1\log _{\mathrm{a}} \mathrm{a}=1
(e) log_((1)/(a))a=-1\log _{\frac{1}{a}} a=-1
(f) log_(a)(x.y)=log_(a)x+log_(a)y;x,y > 0\log _{a}(x . y)=\log _{a} x+\log _{a} y ; x, y>0
(g) log_(a)((x)/(y))=log_(a)x-log_(a)y;x,y > 0\log _{a}\left(\frac{x}{y}\right)=\log _{a} x-\log _{a} y ; x, y>0
(h) log_(a)x^(p)=plog_(a)x;x > 0\log _{a} x^{p}=p \log _{a} x ; x>0
(i) log_(a^(q))x=(1)/(q)log_(a)x;x > 0\log _{a^{q}} x=\frac{1}{q} \log _{a} x ; x>0
(j) log_(a)x=(1)/(log_(x)a);x > 0,x!=1\log _{a} x=\frac{1}{\log _{x} a} ; x>0, x \neq 1
(k) log_(a)x=log_(b)x//log_(b)a;x > 0,a,b > 0,b!=1,a!=1\log _{a} x=\log _{b} x / \log _{b} a ; x>0, a, b>0, b \neq 1, a \neq 1
(l) log_(a)b*log_(b)c*log_(c)d=log_(a)d;a,b,c,d > 0,!=1\log _{a} b \cdot \log _{b} c \cdot \log _{c} d=\log _{a} d ; a, b, c, d>0, \neq 1
(m) a^(log_(a)x)=x;a > 0,a!=1a^{\log _{a} x}=x ; a>0, a \neq 1
(n) a^(log_(b)c)=c^(log_(b)a);a,b,c > 0;b!=1a^{\log _{b} c}=c^{\log _{b} a} ; a, b, c>0 ; b \neq 1
(o) log_(a)x < log_(a)y<=>[[x < y,” if “,a > 1],[x > y,” if “,0 < a < 1]:}\log _{a} x<\log _{a} y \Leftrightarrow\left[\begin{array}{llc}x<y & \text { if } & a>1 \\ x>y & \text { if } & 0<a<1\end{array}\right.
(p) log_(a)x=log_(a)y=>x=y;x,y > 0;a > 0,a!=1\log _{a} \mathrm{x}=\log _{\mathrm{a}} \mathrm{y} \Rightarrow \mathrm{x}=\mathrm{y} ; \mathrm{x}, \mathrm{y}>0 ; \mathrm{a}>0, \mathrm{a} \neq 1
(q) e^(ln a^(x))=a^(x)e^{\ln a^{x}}=a^{x}
(r) log_(10)2=0.3010;log_(10)3=0.4771;ln 2=0.693,ln 10=2.303\log _{10} 2=0.3010 ; \log _{10} 3=0.4771 ; \ln 2=0.693, \ln 10=2.303
(s) If a > 1a>1 then log_(a)x < p=>0 < x < a^(p)\log _{a} x<p \Rightarrow 0<x<a^{p}
(t) If a > 1a>1 then log_(a)x > p=>x > a^(p)\log _{a} x>p \Rightarrow x>a^{p}
(u) If 0 < a < 10<a<1 then log_(a)x < p=>x > a^(p)\log _{a} x<p \Rightarrow x>a^{p}
(v) If 0 < a < 10<a<1 then log_(a)x > p=>0 < x < a^(p)\log _{a} x>p \Rightarrow 0<x<a^{p}