MHR • Advanced Functions 12 Solutions 660

Chapter 7 Section 4 Question 5 Page 392
1
a) log(x 2 ! 3x) 2
= 1 2
1
2 log(x2 ! 3x) = 1 2
log(x 2 ! 3x) = 1
x 2 ! 3x = 10 1
x 2 ! 3x !10 = 0
(x ! 5)(x + 2) = 0
x = 5 or x = !2
Check:
1
a) log(5 2 ! 3" 5) 2
= log10
= 1 2
1
log((!2) 2 ! 3" (!2)) 2
= log10
1
2
1
2
1
b) log(x 2 + 48x) 2
= 1
( ) = 1
1
2 log x2 + 48x
log(x 2 + 48x) = 2
x 2 + 48x = 10 2
x 2 + 48x !100 = 0
(x + 50)(x ! 2) = 0
x = !50 or x = 2
1
b) log((!50) 2 + 48 " (!50)) 2
= log100
= 1
1
log((2) 2 + 48 " 2) 2
= log100
= 1
1
2
1
2
= 1 2
Chapter 7 Section 4 Question 6 Page 392
! x + 5$
a) log 2
"
# 2x %
& = log 256 2
x + 5
2x = 256
x + 5 = 512x
511x = 5
x = 5
511
b) log(2k + 4) ! log k = 1
"
log 2k + 4 %
#
$ k &
' = log10
2k + 4
= 10
k
2k + 4 = 10k
10k ! 2k = 4
8k = 4
k = 1 2
MHR • Advanced Functions 12 Solutions 713