Logarithamic expression concepts-II

By abhijatindia - March 23, 2013

$y = {\log _a}x$ ;

Always remember for base,
a > 0

, $a \ne 1$ (base cant be zero or negative or 1, it can be less than 1 . e.g. 3/2, 1/2)

And for negative value log doesn't exist (why??)
Reason is simple we cant get any negative number even not get zero whatever be the power of a positive number.
e.g. ${(2)^{ - 55}}$ ,
positive or negative what do you think ???
clearly positive.
multiplying 1/2 , 55 times and get positive values.

therefore, x > 0

so, whenever you see a logarithmic expression ,
$y = {\log _a}x$ ; x > 0

and $a \ne 1$

Properties:
1. $lo{g_a}1 = 0$ ; $a > 0,a \ne 1$
clearly, whatever be the value of "a", ${a^0} = 1$

2. ${\log _a}a = 1$ ; $a > 0,a \ne 1$
${a^1} = a$ , $\forall a$

3. $\ln x = {\log _e}x$ and ${{\rm logx}\nolimits} = lo{g_{10}}x$ ; x > 0

4. Base changing formula:-

${\log _a}b = \frac{{\log b}}{{\log a}}$ $= \frac{1}{{\log a/\log b}}$ $= \frac{1}{{{{\log }_b}a}}$ ( in second term you can take any base)

$\Rightarrow {\log _a}b.{\log _b}a = 1$

Proof:
consider, ${\log _a}b$ ; $a > 0,a \ne 1,b > 0$
${\log _a}b = m$ ....................... (i)
$\Rightarrow {a^m} = b$ (by definition)

consider, $\frac{{{{\log }_k}b}}{{{{\log }_k}a}}$ ; $k > 0,k \ne 1$

let ${{{\log }_k}b = q}$ and ${{{\log }_k}a = p}$ ...................... (ii)
$\begin{array}{l} \Rightarrow {k^q} = b\\ {k^p} = a \end{array}$

since, ${a^m} = b$
$\Rightarrow {({k^p})^m} = {k^q}$
$\Rightarrow pm = q$

or, $m = \frac{q}{p}$

therefore, from (i) and (ii),

${\log _a}b = \frac{{\log b}}{{\log a}}$

5. ${\log _c}ab = {\log _c}a + {\log _c}b$ ; $c > 0,c \ne 1$ and a,b > 0

6. $lo{g_c}\frac{a}{b} = {\log _c}a - {\log _c}b$ ; $c > 0,c \ne 1$ and a,b > 0

7. $lo{g_c}{a^b} = b{\log _c}a$ ;    $c > 0,c \ne 1,a > 0$

8. ${\log _{{c^s}}}a = \frac{1}{s}{\log _c}a$ ;    $c > 0,c \ne 1,a > 0$ and $s \ne 0$ (note)

proof is simple left to reader.

9. $a = {b^{{{\log }_b}a}}$

take log with base 'b' both side, you will get the proof.

Examples:

1. Find the value of ${3^{{{\log }_5}7}} - {7^{{{\log }_5}3}}$

$= {7^{{{\log }_7}{3^{{{\log }_5}7}}}} - {7^{{{\log }_5}3}}$

$= {7^{{{\log }_5}7.{{\log }_7}3}} - {7^{{{\log }_5}3}}$

$= {7^{{{\log }_5}3}} - {7^{{{\log }_5}3}}$

= 0

Note ${a^{{{\log }_b}c}} - {c^{{{\log }_b}a}} = 0$

2. (I.I.T)

${\log _{2x + 3}}(6{x^2} + 23x + 21) = 4 - {\log _{3x + 7}}(4{x^2} + 12x + 9)$

for, ${\log _{2x + 3}}(6{x^2} + 23x + 21)$
this logarithmic expression is defined when

$2x + 3 > 0 \Rightarrow x > - 3/2$ , ............... (i)
$2x + 3 \ne 1 \Rightarrow x \ne - 1$ ....................... (ii)

and, $6{x^2} + 23x + 21 > 0$ (quadratic expression here a > 0 and D > 0)

$6{x^2} + 14x + 9x + 21 > 0$

2x(3x + 7) + 3(3x + 7) > 0

or,

from, wavy curvy method,

$x \in ( - \infty , - \frac{7}{3})\bigcup {(\frac{{ - 3}}{2}} ,\infty )$ ........................................ (iii)

for ${\log _{3x + 7}}(4{x^2} + 12x + 9)$
this logarithmic expression is defined when,

$3x + 7 > 0 \Rightarrow x > - 7/3$ ................................................(iv)

$3x + 7 \ne 1 \Rightarrow x \ne - 2$ ....................................................(v)

$4{x^2} + 12x + 9 = {(2x + 3)^2} > 0$

and since square of any number is always greater than or equal to zero

$\Rightarrow x \ne - \frac{3}{2}$ , (since 2x + 3 becomes 0 ) ........................ (vi)

we have to consider only those values of x which satisfy all (i), (ii), (ii), (iv), (v) and (vi)... that is common to all.

because for those values of x , both logarithmic expressions are defined

means x common to all that is intersection of all,
i) x > - 3/2

ii) $x \ne - 1$ ( x belongs to real number except -1)
iii) $x \in ( - \infty , - \frac{7}{3})\bigcup {(\frac{{ - 3}}{2}} ,\infty )$
iv) x > - 7/3

v) $x \ne - 2$ (means x belongs to real number except 2)
vi) $x \ne - \frac{3}{2}$ ( x belongs to real number except -3/2)

clearly intersection of all, that is given equation is defined for ,

$x > - \frac{3}{2},x \ne - 1$ , .............................................(A)

now, ${\log _{2x + 3}}(6{x^2} + 23x + 21) = 4 - {\log _{3x + 7}}(4{x^2} + 12x + 9)$

or, ${\log _{2x + 3}}(2x + 3)(3x + 7) = 4 - {\log _{3x + 7}}{(2x + 3)^2}$

${\log _{2x + 3}}(2x + 3) + {\log _{2x + 3}}(3x + 7) = 4 - 2{\log _{3x + 7}}(2x + 3)$

$1 + {\log _{2x + 3}}(3x + 7) = 4 - 2{\log _{3x + 7}}(2x + 3)$

${\log _{2x + 3}}(3x + 7) = 3 - \frac{2}{{{{\log }_{2x + 3}}(3x + 7)}}$

let $t = {\log _{2x + 3}}(3x + 7)$

$t = 3 - \frac{2}{t}$

${t^2} - 3t + 2 = 0$

$\Rightarrow t = 1$ or t = 2

when

,

${\log _{2x + 3}}(3x + 7) = 1$

$\Rightarrow 3x + 7 = 2x + 3$

$\Rightarrow x = - 4$ ( not in (A) .... i.e. for this value equation is not defined)

when t = 2