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**Unformatted text preview: **nd the zeros of r, we set r(x) = ln(x)+1 = 0 so that
e
e
ln(x) = 0, and we ﬁnd x = e0 = 1. In order to determine test values for r without resorting
to the calculator, we need to ﬁnd numbers between 0, 1 , and 1 which have a base of e. Since
e
1
e ≈ 2.718 > 1, 0 < e1 < 1 < √e < 1 < e. To determine the sign of r e1 , we use the fact that
2
2
e
ln e1 = ln e−2 = −2, and ﬁnd r e1 = −−2 = 2, which is (+). The rest of the test values
2
2
2+1
are determined similarly. From our sign diagram, we ﬁnd the solution to be 0, 1 ∪ [1, ∞).
e
1
Graphing f (x) = ln(x)+1 and g (x) = 1, we see the the graph of f is below the graph of g on
the solution intervals, and that the graphs intersect at x = 1. (+)
0 (−) 0 (+)
1
e 1
y = f (x) = 1
ln(x)+1 and y = g (x) = 1 2. Moving all of the nonzero terms of (log2 (x))2 < 2 log2 (x) + 3 to one side of the inequality,
we have (log2 (x))2 − 2 log2 (x) − 3 < 0. Deﬁning r(x) = (log2 (x))2 − 2 log2 (x) − 3, we get
the domain of r is (0, ∞), due to the presence of the logarithm. To ﬁnd the zeros of r, we
set r(x) = (log2 (x))2 − 2 log2 (x) − 3 = 0 which r...

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