How do you solve #log_5 (4x+3)<log_5 11#?

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Answer 1 · 2017-12-15T16:42:14

Solution : #-0.75 < x < 2 or x| (-0.75 ,2)#

#log_5 (4x+3) < log_5 11#

#4x+3 >0 :. 4x > -3 or x > -3/4 or x > -0.75#

#4x+3 < 11 or 4x <11-3 or 4x <8 #

#or x <2#. Solution : #-0.75 < x < 2#

In interval notation : #x| (-0.75 ,2)#

Answer 2 · 2017-12-15T19:31:56

Graphical interpritation:

I wanted to provide a graphical approach and interpritation:

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