Solve for #x# in #log_3(2x-1) < 0#?

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Thanks in advance

Answer is #(1/2, 1)#

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Answer 1 · 2018-05-09T07:21:34

We know that #log_bx=y# is a continous function defined in #(0,oo)#

We know also that #log_bx# is negative in #(0,1)#

In our case #2x-1# must be smaller than 1 but bigger than 0 (by prior paragraph). Then

#2x-1>0# Then #x>1/2#

By other hand #2x-1<1#, then #2x<2# and so, #x<1#

Combining both results x must be in #(1/2,1)#