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**Unformatted text preview: **esults in a ‘quadratic in disguise.’ We set
u = log2 (x) so our equation becomes u2 − 2u − 3 = 0 which gives us u = −1 and u = 3. Since
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u = log2 (x), we get log2 (x) = −1, which gives us x = 2−1 = 2 , and log2 (x) = 3, which yields
3 = 8. We use test values which are powers of 2: 0 < 1 < 1 < 1 < 8 < 16, and from our
x=2
4
2
sign diagram, we see r(x) < 0 on
is below the graph of y = g (x) = 1
2, 8
2 ln(x)
ln(2) . Geometrically, we see the graph of f (x) =
+ 3 on the solution interval. ln(x)
ln(2) 2 372 Exponential and Logarithmic Functions (+) 0 (−) 0 (+)
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2 0 8
y = f (x) = (log2 (x))2 and y = g (x) = 2 log2 (x) + 3 3. We begin to solve x log(x +1) ≥ x by subtracting x from both sides to get x log(x +1) − x ≥ 0.
We deﬁne r(x) = x log(x +1) − x and due to the presence of the logarithm, we require x +1 > 0,
or x > −1. To ﬁnd the zeros of r, we set r(x) = x log(x + 1) − x = 0. Factoring, we get
x (log(x + 1) − 1) = 0, which gives x = 0 or log(x +1) −...

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