SOLUTION: What values of X cannot possibly be solutions of the following equation:
log a (3x+1)=2
(note: the "a" is suppossed to be the base of the logarithm)
I can not find this
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Question 20954: What values of X cannot possibly be solutions of the following equation:
log a (3x+1)=2
(note: the "a" is suppossed to be the base of the logarithm)
I can not find this in the book.
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
log a (3x+1)=2
USE FORMULA LOG xTO BASE Y = LOG X/LOG Y...ALL LOGS TAKEN WITH A COMMON BASE
log a (3x+1)=2=LOG (3X+1)/LOG A =2
LOG (3X+1)=2*LOG A ..WE HAVE TO FIND THE RESTRICTIONS ON X.
SO WE TAKE THAT 2*LOG A IS SOME REAL NUMBER (OF COURSE FOR THIS TO HAPPEN A ALSO HAS TO BE A POSITIVE NUMBER).FOR THAT TO HAPPEN LOG (3X+1) HAS TO BE REAL NUMBER
SO 3X+1>0,SINCE LOG 0 OR LOG OF NEGATIVE NUMBER DOES NOT EXIST.HENCE
3X+1>0...OR
3X>-1...OR..
X>-1/3
HENCE X HAS TO BE GREATER THAN -1/3 FOR THE ABOVE EQUATION TO HOLD TRUE FOR ANY REAL VALUE OF A >0
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