Logarithm Definition and Properties.
Logarithm and properties of logarithmic function was used as a calculating tool before calculator were discovered.
Definition of Logarithm is the index m such that $a^m=n$ where a>0 , $a\neq 1$ and n > 0.
i.e If $a^m=n$ then the index m is defined as $m=log_a(n)$ (m is logarithm of n to the base a).
Vise-Versa If $m=log_a(n)$ then $n=a^m$.
If function $y=f(x)=a^x$ where a>0 and $a\neq 1$ then $x=log_a(y)$.
Now replace x$\rightarrow $ y and y$\rightarrow$ x in $x=log_a(y)$ we get $y=log_a(x)$ where a>0 and $a\neq 1$.
Hence a logarithmic function is defined as $y=g(x)=log_a(x)$ if and only if $x=a^y$ where a>0 and $a\neq 1$ and $x\in (0,\infty)$.
Domain and Range of logarithmic function $f(x)=log_a(x)$ is
Domain of f(x) is $x\in (0,\infty)$ AND Range of of f(x) is $f(x)\in R$
Properties of Logarithmic function.
Logarithm and properties which is used to evaluate/ simplify various problems involving logarithmic functions. These properties are listed bellow.
(1) Since $a^0=1$ Hence $log_a(1)=0$ for a > 0 and $a\neq 1$.
(2) We know that $a^1=a$ then $log_a(a)=1$ where a>0 and $a\neq 1$.
(3) If $log_a(x)=y$ then $x=a^y$ where a>0 and $a\neq 1$.
(4) $log_a(x_1)=log_a(x_2)$ if only if $x_1=x_2$ where $x_1,x_2\in (0,\infty)$
(5) $log_a(m) + log_a(n)=log_a(mn)$ where a>0 , $a\neq 1$ and n,m>0.
Proof: Let $log_a(m)=x$ and $log_a(n)=y$ for some a>0 and $a\neq 1$
$\Rightarrow \;\; m=a^x$ and $n=a^y$
$\therefore \;\; mn=(a^x)(a^y)$
$\Rightarrow \; mn=a^{(x+y)}$
$\Rightarrow \;\; x+y=log_a(mn)$
Now Replace x and y with $log_a(m)$ and $log_a(n)$ respectively.
Hence $log_a(m) + log_a(n) = log_a(mn)$
(6) $log_a(m) – log_a(n) = log_a(\cfrac{m}{n})$
Proof: Let $log_a(m)=x$ and $log_a(n)=y$ for some a>0 and $a\neq 1$
$\Rightarrow \;\; m=a^x$ and $n=a^y$
$\therefore \;\; (\cfrac{m}{n})=\cfrac{(a^x)}{(a^y)}$
$\Rightarrow \; (\cfrac{m}{n})=a^{(x-y)}$
$\Rightarrow \;\; x-y=log_a(\cfrac{m}{n})$
Now Replace x and y with $log_a(m)$ and $log_a(n)$ respectively.
Hence $log_a(m) – log_a(n) = log_a(\cfrac{m}{n})$
(7) Base Change Property $log_a(m)=\cfrac{log_b(m)}{log_b(a)}$ Where b>0 and $b\neq 1$
(a) If new base is made as m where $m\neq 1$ then $log_a(m)=\cfrac{log_m(m)}{log_m(a)} $
$\Rightarrow \;\; log_a(m)=\cfrac{1}{log_m(a)}$.
Hence $log_a(m).log_m(a)=1$
(b) $log_{x_1}x_2.log_{x_2}x_3.log_{x_3}x_4\ldots log_{x_{(n-1)}}{x_n}$ = $log_{x_1}{x_n}$
(8) $a^{(log_a(m))}$=m
Proof: Let $log_a(m)=x$
$\Rightarrow \; m=a^x$ Put $x=log_a(m)$
Hence $m=a^{log_a(m)}$
(9) $a^{(log_b(m))}$ = $m^{(log_b(a))}$
Proof: Let $a^{log_b(m)}$ = $a^{\cfrac{log_a(m)}{log_a(b)}}$
Using $log_a(b)=\cfrac{1}{log_b(a)}$
We get , $a^{log_b(m)}$ = $a^{(log_b(a))(log_a(m))}$
Using $nlog_a(m)=log_a(m^n)$
We get , $a^{log_b(m)}$ = $ a^{log_a(m^{log_b(a)})}$
Using $a^{log_a(x)}=x$
We get , $a^{log_b(m)}$ = $m^{log_b(a)}$
Logarithmic function and Inequalities.
For f(x)=$log_a(x)$ where 0<a<1
(1) If $0\lt x_1 \lt x_2$ then $log_a(x_1)\gt log_a(x_2)$.
(2) If $log_a(x_1) \gt log_a(x_2)$ then $0\lt x_1 \lt x_2$
(3) If $log_a(x)\lt n$ then $x\gt a^n$.
(4) If $log_a(x)\gt n$ then $0\lt x\lt a^m$
(5) If $n_1\lt log_a(x)\lt n_2$ then $a^{n_2} \lt x \lt a^{n_1}$
For f(x)=$log_a(x)$ where a>1
(1) If $0\lt x_1 \lt x_2$ then $log_a(x_1)\lt log_a(x_2)$.
(2) If $log_a(x_1) \lt log_a(x_2)$ then $0\lt x_1 \lt x_2$
(3) If $log_a(x)\lt n$ then $0\lt x\lt a^n$.
(4) If $log_a(x)\gt n$ then $x\gt a^m$
(5) If $n_1\lt log_a(x)\lt n_2$ then $a^{n_1} \lt x \lt a^{n_2}$
Logarithm Practice Questions.
Example 1 :: Find $x$ such that
(a) $log_48 + log_4(x+3)-log_4(x-1)=2$
(b) $log_{10}(-x)= 2log_{10}(x+1)$
Example 1 :: Find $x$ such that
(a) $log_48 + log_4(x+3)$ $-log_4(x-1)=2$
(b) $log_{10}(-x)= 2log_{10}(x+1)$
Solution 1 ::
(a) Given $log_48 + log_4(x+3)$ $ – log_4(x-1)=2$
$\Rightarrow$ $log_4(\cfrac{8(x+3)}{x-1}) = 2 $
$\Rightarrow$ $\cfrac{8(x+3)}{x-1}=4^2=16$
$\Rightarrow$ $\cancel{8} (\cfrac{x+3}{x-1})=\cancel{16} \,2$
$\Rightarrow$ $x+3 = 2 (x-1 )$
$\Rightarrow$ $x+3 = 2x -2$
$\therefore$ $x=5$ For $x=5$ Both (x+3) and (x-1) are positive hence all log terms are well defined.
Hence x=5 is the solution
(b) For $log_{10}(-x)$ to exist $-x\gt 0$ hence $x\lt 0$ and for $log_{10}(x+1)$ to exist $x\gt -1$.
Hence For given equation to exist $-1\lt x\lt 0$
Now for equation, $log_{10}(-x)= 2log_{10}(x+1)$
$\Rightarrow$ $log_{10}(-x) = log_{10}(x+1)^2$
$\Rightarrow$ $-x=(x+1)^2$
$\Rightarrow$ $x^2+3x+1=0$
$\therefore $ $x=\cfrac{-3\pm \sqrt{5}}{2}$
$\therefore $ $x=\cfrac{\sqrt{5} – 3}{2}$ is the solution.
Example 2 :: Find $x$ such that $(0.1)^{4x^2-2x-2} \leq (10)^{3-2x}$
Solution 2 :: Given $(0.1)^{4x^2-2x-2}\leq (10)^{3-2x}$
$\Rightarrow$ $(0.1)^{4x^2-2x-2}\leq (0.1)^{2x-3}$
$\Rightarrow$ $4x^2-2x-2 \geq 2x-3 $
$\Rightarrow$ $4x^2-4x+1 \geq 0$
$\Rightarrow$ $(2x^2-1)^2 \geq 0$
$\Rightarrow$ $x\in R$
Hence Solution is $x \in R$
Example 3 :: Find $x$ such that $1+2log_{(x+2)}5=log_5(x+2)$
Solution 3 :: Given $1 + 2log_{(x+2)}5 = log_5(x+2)$
Let $log_5(x+2) = y$ $\therefore \; log_{(x+2)}5=\cfrac{1}{y} $
Put in given equation, hence
$1 + 2 \cfrac{1}{y} = y$
$\Rightarrow$ $y +2 = y^2$
$\Rightarrow$ $y^2 -y -2 = 0$
$\Rightarrow$ $(y-2)(y+1)=0$
$\therefore$ $y=log_5(x+2) = -1 ; 2 $
$\therefore$ $x+2 = 5^{-1}\,;\, 5^2$
$\therefore $ $x=\cfrac{-9}{5}\,;\, 23 $
Hence Solution is $x= \cfrac{-9}{5}\,;\, 23 $