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**Unformatted text preview: **ter, we isolate the exponential and take logs to get 2x = ln(4), or x = ln(4) = ln(2). (Can
2
you explain the last equality using properties of logs?) As in the previous example, we need
to be careful about choosing test values. Since ln(1) = 0, we choose ln 1 , ln 3 and ln(3).
2
2
1
Evaluating,6 we have r(ln 1 ) = ln 1 e2 ln( 2 ) − 4 ln 1 . Applying the Power Rule to the log
2 2
12
ln( 2 ) 2 1
1
in the exponent, we obtain ln
e
− 4 ln
= ln 1 eln( 4 ) − 4 ln 2 . Using the inverse
2
1
1
1
15
1
1
properties of logs, this reduces to 4 ln 2 − 4 ln 2 = − 4 ln 2 . Since 1 < 1, ln 2 < 0
2
and we get r(ln 1 ) is (+). Continuing in this manner, we ﬁnd r(x) < 0 on (0, ln(2)). The
2
calculator conﬁrms that the graph of f (x) = xe2x is below the graph of g (x) = 4x on this
intervals.7 1
2 1
2 (+) 0 (−) 0 (+)
0 ln(2)
y = f (x) = xe2x and
y = g (x) = 4x 6
7 A calculator can be used at this point. As usual, we proceed without apologies, with the analytical method.
Note: ln(2) ≈ 0.693. 364 Exponential and Logarithmic Functions Example 6.3.3. Re...

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