MHR • Advanced Functions 12 Solutions 660
MHR • Advanced Functions 12 Solutions 660
MHR • Advanced Functions 12 Solutions 660
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Chapter 7 Review Question 7 Page 408<br />
a) (4 x ) 2 – 4 x – 20 = 0<br />
a = 1, b = –1, c = –20<br />
4 x = !(!1) ± (!1)2 ! 4(1)(!20)<br />
2(1)<br />
4 x =<br />
1± 81<br />
2<br />
4 x = 1± 9<br />
2<br />
4 x = 5 or 4 x = !4 extraneous root since the base is negative<br />
x log 4 = log5<br />
x = log5<br />
log 4<br />
b) 2 x (2 x ) +<strong>12</strong>(2 ! x )(2 x ) ! 7(2 x ) = 0<br />
( 2 ) x 2 ! 7 2 x<br />
( ) +<strong>12</strong> = 0<br />
(2 x ! 4)(2 x ! 3) = 0<br />
2 x = 4 or 2 x = 3<br />
2 x = 2 2 x log 2 = log3<br />
x = 2<br />
x = log3<br />
log 2<br />
Chapter 7 Review Question 8 Page 408<br />
!<br />
a) 1516 = 2000 1 $<br />
"<br />
# 2%<br />
&<br />
0.758 = 1 2<br />
1<br />
h<br />
log0.758 = 1 h log 1 2<br />
1<br />
h = log0.758<br />
log 1 2<br />
h =<br />
log 1 2<br />
log0.758<br />
h &= 2.5<br />
The half-life is approximately 2.5 years.<br />
1<br />
h<br />
t<br />
2.5<br />
b) 0.10 &= 1 2<br />
log0.10 &=<br />
t<br />
2.5 log 1 2<br />
t log0.10<br />
&=<br />
2.5<br />
log 1 2<br />
t &= 2.5log0.10<br />
t &= 8.3<br />
log 1 2<br />
It will take approximately 8.3 years to<br />
depreciate to 10% of its purchase price.<br />
<strong>MHR</strong> <strong>•</strong> <strong>Advanced</strong> <strong>Functions</strong> <strong>12</strong> <strong>Solutions</strong> 738