Logarithmic expression inequality concepts - III
MOST IMPORTANT CONCEPTS
if,
,
, 
i) If
, 
ii) If
, 
why ?????
take

increasing with increase in power,

log is also increasing, with increase in the value of 'x' (
)
means,

and
means,
is increasing with increase in the value of "x", if 
Now, take
decreasing with increase in power

decreasing with increase in value of x (
)
i.e.

i.e.
is decreasing with increase in value of "x" if 
Thus,
if,
,
, 
i) If
, 
ii) If
, 
By Graph,
we know,
clearly from figure,
if,
,
, 
i) If
, 
ii) If
, 
Example:

first find the values of x for which the inequality is defined.
STEP 1:
(since
,
)
or,
by wavy curvy, (previous post),
........................................... (A)
therefore inequality is defined in above interval,
STEP 2:
now solve the inequality,

since base is greater than 1, log is increasing, we know for any base ,
( we can also see from graph of
, that for x > 1, value of
is grater than zero, therefore for
,
)



from wavy-curvy method,
.......................... (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
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
if,
i) If
ii) If
why ?????
take
increasing with increase in power,
log is also increasing, with increase in the value of 'x' (
means,
and
means,
Now, take
decreasing with increase in power
decreasing with increase in value of x (
i.e.
i.e.
Thus,
if,
i) If
ii) If
By Graph,
we know,
clearly from figure,
if,
i) If
ii) If
Example:
first find the values of x for which the inequality is defined.
STEP 1:
(since
or,
by wavy curvy, (previous post),
therefore inequality is defined in above interval,
STEP 2:
now solve the inequality,
since base is greater than 1, log is increasing, we know for any base ,
from wavy-curvy method,
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
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
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