Logarithmic expression inequality concepts - III

MOST IMPORTANT CONCEPTS

if,  {\log _c}a > {\log _c}b  , c > 0,c \ne 1 , a,b > 0

i) If c > 1 ,   a > b

ii) If c < 1  , a < b

why ?????

take  c = 2,c > 1

\begin{array}{l}
{2^2} = 4,\\
{2^3} = 8\\
{2^4} = 16
\end{array}

increasing with increase in power,

\begin{array}{l}
{\log _2}4 = 2\\
{\log _2}8 = 3\\
{\log _2}16 = 4
\end{array}

log is also increasing, with increase in the value of 'x' ({\log _c}x,c > 1)
means,
4 < 8 < 16

and {\log _2}4 < {\log _2}8 < {\log _2}16

means, y = {\log _c}x is increasing with increase in the value of  "x",  if c > 1

Now, take c = 1/2

\begin{array}{l}
{\left( {\frac{1}{2}} \right)^2} = \frac{1}{4}\\
{\left( {\frac{1}{2}} \right)^3} = \frac{1}{8}\\
{\left( {\frac{1}{2}} \right)^4} = \frac{1}{{16}}
\end{array}

decreasing with increase in power

\begin{array}{l}
{\log _{\frac{1}{2}}}\frac{1}{4} = 2\\
{\log _{\frac{1}{2}}}\frac{1}{8} = 3\\
{\log _{\frac{1}{2}}}\frac{1}{{16}} = 4
\end{array}

decreasing with increase in value of x ({\log _c}x,c < 1 )

i.e. 1/16 < 1/8 < 1/4

{\log _{\frac{1}{2}}}\frac{1}{{16}} > {\log _{\frac{1}{2}}}\frac{1}{8} > {\log _{\frac{1}{2}}}\frac{1}{4}

i.e. y = {\log _c}x is decreasing with increase in value of "x" if c < 1

Thus,


if,  {\log _c}a > {\log _c}b  , c > 0,c \ne 1 , a,b > 0

i) If c > 1 ,   a > b

ii) If c < 1  , a < b


By Graph,

we know,{\log _c}1 = 0



clearly from figure, 


if,  {\log _c}a > {\log _c}b  , c > 0,c \ne 1 , a,b > 0

i) If c > 1 ,   a > b

ii) If c < 1  , a < b


Example:

{\log _{\frac{3}{2}}}\left( {\frac{{2x - 8}}{{x - 2}}} \right) > 0


first find the values of x for which the inequality is defined.

STEP 1:

\left( {\frac{{2x - 8}}{{x - 2}}} \right) > 0

(since {\rm{denominator}} \ne 0,  x \ne 2)

or, \frac{{x - 4}}{{x - 2}} > 0
by wavy curvy, (previous post),

x \in ( - \infty ,2)\bigcup {(4,\infty )}   ........................................... (A)

therefore inequality is defined in above interval,



STEP 2:
now solve the inequality,

{\log _{\frac{3}{2}}}\left( {\frac{{2x - 8}}{{x - 2}}} \right) > 0

since base is greater than 1, log is increasing, we know for any base , {\log _c}1 = 0

{\log _{\frac{3}{2}}}\left( {\frac{{2x - 8}}{{x - 2}}} \right) > {\log _{\frac{3}{2}}}1 

 \Rightarrow \left( {\frac{{2x - 8}}{{x - 2}}} \right) > 1 ( we can also see from graph of y = {\log _c}x,c > 1, that for x > 1, value of y = {\log _c}x is grater than zero, therefore for  {\log _{\frac{3}{2}}}\left( {\frac{{2x - 8}}{{x - 2}}} \right) > 0 ,\left( {\frac{{2x - 8}}{{x - 2}}} \right) > 1 )




\frac{{2x - 8}}{{x - 2}} - 1 > 0

\frac{{2x - 8 - x + 2}}{{x - 2}} > 0

\frac{{x - 6}}{{x - 2}} > 0

from wavy-curvy method,


x \in ( - \infty ,2)\bigcup {(6,\infty )}  .......................... (B)

STEP 3:

but the inequality is defined on (A)

therefore, only those solutions are valid or defined which lies in the interval (A)

take the intersection of (A) and (B) ,

we get the solution of inequality,

ANSWER x \in ( - \infty ,2)\bigcup {(6,\infty )}  


RULES/STEPS TO SOLVE INEQUALITY/EQUALITY


STEP 1. First find the values for which inequality/equality is defined (Let the solution be "A")

STEP 2. Solve the inequality/equality ( Let the solution is "B")

STEP 3. After solving the inequality take only those values from the solution which belong to "A" ( STEP 1 INTERVAL/SETS).

                                                          OR

Take the intersection of STEP 1 interval/sets and STEP 2 interval/sets that is A\bigcap B













Comments

Popular posts from this blog

Properties of Modulus & Argument: Complex Number

Fractional Part function

Geometrical Meaning of Argument and Modulus - I: Complex Number