How do I solve log_3(2x-1)≤2 for x?

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Answer 1 · 2018-05-09T03:47:37

#1/2 < x ≤ 5# will be our solution

You're going to want to solve for #x#.

#2x - 1 ≤ 3^2#

#2x - 1 ≤ 9#

#2x ≤ 10#

#x ≤ 5#

However, we need #2x - 1 > 0#, or #x > 1/2#. Therefore, our answer is

#1/2 < x ≤ 5#

We may confirm graphically.

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Hopefully this helps!