082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

86 Chapter 2 Engineering functions
Solution (a) 50 = 9(10 2x ) (b) 3e −(2x+1) = 10
10 2x = 50 e −(2x+1) = 10
9
2x = log 50 3
−(2x +1) = ln 10 9
x = 1 2 log 50 3
9 = 0.372 2x=−ln 10 3 −1
(c) log(x 2 −1) = 2
x 2 −1=10 2 =100
x 2 = 101
x = ±10.050
2x = −2.204
x = −1.102
(d) 3ln(4x +7) = 12
4x+7=e 4
4x=e 4 −7
x = e4 −7
= 11.900
4
Logarithmicexpressionscanbemanipulatedusingthelawsoflogarithms.Theselaws
are identical for any base, but it is essential when applying the laws that bases are not
mixed.
log a
A +log a
B = log a
(AB)
( ) A
log a
A −log a
B = log a
B
nlog a
A = log a
(A n )
log a
a=1
Wesometimesneedtochangefromonebasetoanother.Thiscanbeachievedusingthe
following rule.
Inparticular,
Example2.14 Simplify
log a
X = log b X
log b
a
log 2
X = log 10 X
log 10
2 = log 10 X
0.3010
(a) logx +logx 3
(b) 3logx +logx 2
( 1
(c) 5lnx+ln
x)
(d) log(xy) +logx −2logy
( ) 4
(e) ln(2x 3 ) −ln + 1 x 2 3 ln27
Solution (a) Usingthe laws of logarithms we find
logx +logx 3 = logx 4