Login

How had the expression become equal to 1? (Please view the attatched photo).

How had the expression become equal to 1? (Please view the attatched photo). logx+logxx2 1+10gxx EXAMPLE 2 lfa, b, care positive real numbers, then x dog a-log b none of these (d) Ans. (b) SOLUTION We have, log blogc-loga cloga-logb = (log b- log c) log a + (log c — log a) log b + (log a — log b) log c alogb-logc blogc—loga clog a—logb = 1 are positive real numbers, then the value of (ab) x (bc) x (ca) (c) EXAMPLE I xyz +1 (a) 2yz Ans. (a) SOLUTION xyz + xyz + xyz -4 xyz - xyz EXAMPLI 10 Ans. (b

Please find below the solution to the asked query:

From given solution we consider step :

log { alog b - log c. blog c - log a . clog a - log b } = 0  ( As you have doubt in next step )

Now we take " Antilog " and get :

alog b - log c. blog c - log a . clog a - log b = Antilog 0

alog b - log c. blog c - log a . clog a - log b = (10)0               ( We know : Antilog x  =  10x   )

alog b - log c. blog c - log a . clog a - log b = 1                     ( We know ( Anything )0 = 1 )


Hope this information will clear your doubts about topic.

  • 2
View Full Answer
ACCORDING TO ONE PROPERTY OF LOG
IF BASE IS NOT MENTIONED THEY ARE CONSIDERED TO BE SAME 
SO LOGa &  LOGb & LOGc will have same base now acc to identity                  1. logm/n=logm-logn  (keeping base same)
USING INVERSE OF THIS PROPERTY
logb-logc=log(b/c)
logc-loga=log(c/a)
loga-logb=log(a/b)
ok                                                    loga^x 
now acc to identity of log               ( a              = x )  now keeping base =constant

a^logb/c= b/c
similarly 
b^logc/a= c/a
&
c^loga/b= a/b
now b/c * c/a * a/b =1

hence , prooved
  • 1
What are you looking for?