Exponential functions 125

11. 5= 8

(

1 −e

−x

2

)

12. ln(x+ 3 )−lnx=ln(x− 1 )


13. ln(x− 1 )^2 −ln3=ln(x− 1 )


14. ln(x+ 3 )+ 2 = 12 −ln(x− 2 )


15. e(x+^1 )= 3 e(^2 x−^5 )


16. ln(x+ 1 )^2 = 1. 5 −ln(x− 2 )+ln(x+ 1 )


17. Transpose b=lnt−alnD to make t the
    subject.


18. If

P

Q

=10log 10

(

R 1

R 2

)

, find the value ofR 1

whenP=160,Q=8andR 2 =5.

19. IfU 2 =U 1 e

(W

PV

)

, make W the subject of the

formula.

16.5 Laws of growth and decay

Laws of exponential growth and decay are of the form
y=Ae−kxandy=A( 1 −e−kx), whereAandkare
constants. When plotted, the form of these equations is
as shown in Figure 16.5.

y

A

0

y 5 Ae^2 kx

y 5 A(1 2 e^2 kx^ )

x

y

A

(^0) x
Figure 16.5
The laws occur frequently in engineering and science
and examples of quantities related by a natural law
include:
(a) Linear expansion l=l 0 eαθ
(b) Change in electrical resistance with temperature
Rθ=R 0 eαθ
(c) Tension in belts T 1 =T 0 eμθ
(d) Newton’s law of cooling θ=θ 0 e−kt
(e) Biological growth y=y 0 ekt
(f) Discharge of a capacitor q=Qe−t/CR
(g) Atmospheric pressure p=p 0 e−h/c
(h) Radioactive decay N=N 0 e−λt
(i) Decay of current in an inductive circuit
i=Ie−Rt/L
(j) Growth of current in a capacitive circuit
i=I( 1 −e−t/CR)
Here are some worked problems to demonstrate the laws
of growth and decay.
Problem 18. The resistanceRof an electrical
conductor at temperatureθ◦Cisgivenby
R=R 0 eαθ,whereαis a constant andR 0 =5k.
Determine the value ofαcorrect to 4 significant
figures whenR=6kandθ= 1500 ◦C. Also, find
the temperature, correct to the nearest degree, when
the resistanceRis 5.4k
TransposingR=R 0 eαθgives
R
R 0
=eαθ
Taking Napierian logarithms of both sides gives
ln
R
R 0
=lneαθ=αθ
Hence,α=
1
θ
ln
R
R 0

1
1500
ln
(
6 × 103
5 × 103
)

1
1500
( 0. 1823215 ...)
= 1. 215477 ...× 10 −^4
Hence,α=1.215× 10 −^4 correct to 4 significant
figures.
From above, ln
R
R 0
=αθhenceθ=
1
α
ln
R
R 0