ok. let's solve ln x <0
solve ln x = 0
raise e to both sides
e^lnx = e^0
x=1
so we make a number line:
we add this root and any place that ln is undefined. you can't take the ln of anything less than or equal to 0, so we also put 0.
-undef---------------neg----------------...
.............0.................1...........
we test a number bigger than 1 and a number less than 1.
since e>1, we can use it, and ln e = ln(e^1)= 1 > 0
since 0<1/e < 1, we can use it, and ln(1/e) = ln(e^-1) = -1 < 0
so ln x < 0 when 0
now lets try e^x > 6.
solve e^x = 6.
take ln of both sides.
x = ln 6.
------ < 6-------------------------->6-----------...
....................ln 6...............................
e^x is always defined so we test a number bigger than ln 6 and a number less than ln 6.
0 < ln 6, so we can use it and e^0 = 1 < 6.
ln 7 > ln 6, since ln is always increasing. so we can use ln7, and e^ln 7 = 7 > 6.
so e^x > 6 for x>ln 6 or (ln 6, infinity).
Hope it helps