How do you solve #log_5(2x-1) < 1# ?

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Answer 1 · 2015-06-21T09:28:46

The answer is #x<3# or # x in (-oo;3)#

#log_5(2x-1) <1#

First we have to change the number on the right side to a logarythm using the rule that

#a=log_b(b^a)#

So we get:

#log_5(2x-1) < log_5(5)#

Now, when we have logarythm on both sides of the inequality and the base is the same we can leave the logarythm

#2x-1<5#

Note, that if the base #b# was in the range #(0;1)# we would have to change the sign of the inequality from #<# to #>#.

Now we have to solve this linear inequality:
#2x<6#

#x<3# or #x in (-oo;3)#

Answer 2 · 2015-06-21T09:29:40

Use properties of #log# and exponentiation to find:
#1/2 < x < 3#

Firstly, #log_5(2x-1)# is only defined when #2x-1 > 0#, so #x > 1/2#.

Secondly, #log_5:(0,oo) -> RR# is a monotonically increasing function with inverse #5^x:RR -> (0, oo)#. So we can deduce:

#5^(log_5(2x-1)) < 5^1#

That is, #2x-1 < 5#.

Add #1# to both sides, then divide both sides by #2# to get

#x < 3#

Putting these two conditions together we get:

#1/2 < x < 3#