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**Unformatted text preview: **to divide by ‘ln’. 6.3 Exponential Equations and Inequalities 359
1−x 1. Since 16 is a power of 2, we can rewrite 23x = 161−x as 23x = 24
. Using properties of
exponents, we get 23x = 24(1−x) . Using the one-to-one property of exponential functions, we
4
get 3x = 4(1 − x) which gives x = 7 . To check graphically, we set f (x) = 23x and g (x) = 161−x
and see that they intersect at x = 4 ≈ 0.5714.
7
2. We begin solving 2000 = 1000 · 3−0.1t by dividing both sides by 1000 to isolate the exponential
which yields 3−0.1t = 2. Since it is inconvenient to write 2 as a power of 3, we use the natural
log to get ln 3−0.1t = ln(2). Using the Power Rule, we get −0.1t ln(3) = ln(2), so we
divide both sides by −0.1 ln(3) to get t = − 0.ln(2) = − 10 ln(2) . On the calculator, we graph
1 ln(3)
ln(3)
f (x) = 2000 and g (x) = 1000 · 3−0.1x and ﬁnd that they intersect at x = − 10 ln(2) ≈ −6.3093.
ln(3) y = f (x) = 23x and
y = g (x) = 161−x y = f (x) = 2000 and
y = g (x) = 1000...

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