CHAPTER
4
SOLUTIONS
TO
REINFORCEMENT
EXERCISES
IN
EXPONENTIALS AND LOGARITHMS
4.3.1 y = an n = an integer
4.3.1A.
Plot the values of 3n for n = – 2, – 1, 0, 1, 2, using Cartesian axes with n on the
horizontal axis and 3n on the vertical axis.
Solution
This should be straightforward for you now - the form of the graph is shown
in the solution to B.
B.
Plot the graphs of y = 3x and y = 4x on the same axes. Sketch the graph of
y = x.
Solution
Note that since 3 <  < 4, the graph of y = x lies sandwiched between those
of y = 3x and y = 4x as shown.
4.3.2 The general exponential function ax
Simplify
8x  23x
6x/2 12x+1 27–x/2
i)
ii)
43x
32x/2
iii)
x –1/3y–2/3
(x2 y4)–1/6
x2(x2 + 1)–1/2 – (x2 + 1)1/2
iv)
x2
-1-
v)
vii)
ex e–x
2
ex–1 e(x+1)
vi)
2
a3 a–(x+1)
2
2
a–x a–2x
acos2x a–cos2x
2
a3–sin x
Solution
i) It is best to first express everything in terms of powers of 2, then:
8x  23x
43x
(23)x  23x
23x  23x 26x
=
=
= 6x = 1
(22)3x
26x
2
ii) In this case we can express everything in terms of powers of 2 and 3:
6x/2 12x+1 27–x/2
(2  3x/2  (22  3)x+1  (33)–x/2
=
32x/2
(25)x/2
Now collect terms:
= 2x/2 22(x + 1) 2-5x/2 3x/2 3x + 1 3–3x/2
= 22x + 2 + x/2 – 5x/2 3x +
1 + x/2 – 3x/2
= 22 3 = 12
So in this case the x dependence cancels out.
iii)
x –1/3y–2/3
(x2 y4)–1/6
=
x –1/3y–2/3
(x2)–1/6(y4)–1/6
=
x –1/3y–2/3
=1
x–1/3y–2/3
iv) In this case, multiply top and bottom by (x2 + 1)1/2 to obtain
x2(x2 + 1)–1/2 – (x2 + 1)1/2
x2
=
x2 – (x2 + 1)
= 2 2
x (x + 1)1/2
–1
1
1/2 = – 2
x (x + 1)
x x2 + 1
2
2
-2-
v)
ex e–x
2
ex–1 e(x+1)
2
ex e–x
=
ex–1 ex
2
= ex e–x e– x+ 1 e– x
2
2+2 x+1
2–2x–1
2
= e– 2x e– 2x
= e– 2x(x + 1)
vi)
a3 a–(x+1)
2
2
a–x a–2x
= a3 a– x
=
a3 a– x
2 – 2x - 1 x2 2x
a a
2 – 2x – 1
2
a–x a–2x
= a3– x
2 – 2x – 1+ x2 + 2x
= a2
vii) Remember the trig identities cos 2x = 2 cos2x – 1 and cos2x + sin2x = 1:
acos2x a– cos2x
a2cos2x – 1 a– cos2x
=
2
2
a3 – sin x
a3– sin x
2
2
= a2cos2x – 1 a– cos2xa– 3 + sin x = a2cos2x – 1 a– cos2xa– 3 + sin x
= a2cos2x – 1 – cos2x – 3 + sin2x = a – 3
4.3.3 The natural exponential function ex
4.3.3A.
Referring to the 'interesting' problem in Section 4.2.3, determine the debt owing if
interest is reckoned by the i) minute
ii) second.
Solution
i) There are 8760  60 = 525600 minutes in a non-leap year, so the debt
reckoned by the minute is
I

 525600
£ C1 + 52560000


at the end of the year.
ii) There are 525600  60 = 31536000 minutes in a non-leap year, so the debt
reckoned by the second is
-3-
I

 31536000
£ C1 + 3153600000


at the end of the year.
4.3.3B.
Use the series for ex to evaluate to 4 decimal places i) e
ii) e0.1
iii) e2
Solution
The series for the exponential function is
x2 x3
xr
e = 1 + x + 2! + 3! + … + r! + …
x
i) For x = 1 we therefore have
1
1
1
1
e = 1 + 1 + 2! + 3! + 4! + … + r! + …
1
1
1
= 1 + 1 + 2! + 3! + 4!
+ ...
= 1 + 1 + 0.5 + 0.16667 + 0.04167 + 0.00833 + 0.00139 + 0.0002 + ...
= 2.7183 to four decimal places.
ii) For x = 0.1 we have
e
0.1
(0.1)2
(0.1)3
(0.1)4
(0.1)r
= e = 1 + 0.1 + 2!
+ 3!
+ 4!
+ … + r!
+ …
= 1 + 0.1 + 0.005 + 0.00017 + 0.000004 ...
= 1.1052 to 4 dp
iii) For x = 2 we have
22
23
24
2r
e2 = e = 1 + 2 + 2! + 3! + 4! + … + r! + …
-4-
= 1 + 2 + ...
= 7.3891 to 4 dp
4.3.3C.
Sketch the curves
i)
y = ex – 1
y = 1 – e–x
ii)
Solution
i)
ii)
4.3.3D.Solve the equation
e2x – 2ex + 1 = 0
Solution
If we put u = ex then the equation becomes
(ex)2 – 2ex + 1 = u2 – 2u + 1 = (u –1)2 = 0
giving u = ex = 1 and so
x=0
4.3.4 Manipulation of the exponential function.
Simplify
i)
A
B 2 3C
e (e ) e
ex e3y(ex)
ii)
e–x ey
2
-5-
iii)
(e x )3 e x 2 ( e y )2
e4y ex+3
(eB)2 eA eB–2
iv)
v)
e2–B e2A
(eA + e–A)
2
vi)
e2A
(eA + e3B)(e–A + e–B)
e2B eA
Solution
2
i) eA (eB) e3C = eA e2B e3C = eA +
2B + 3C
2
ex e3y(ex)
= ex e3ye2xex e– y = ex e2xexe3ye– y = e4x e2y = e2(2x + y)
e–x ey
ii)
iii)
(e x )3 e x 2 (e y )2
e4y ex+3
2
e3x ex e2y
2
2
= 4y x+3 = e3x ex e2ye– 4y e– x– 3 = e3x + x + 2y – 4y - x - 3
e e
= ex
(eB)2 eA eB–2
iv)
e2–B e2A
2
+ 2x – 2y - 3
e2B eA eB–2
= e2B eA eB–2eB – 2 e– 2A
e2–B e2A
=
= e– A + 4B – 4
v)
(eA + e–A)
e2A
2
2
=
(eA) + 2eA e–A + (e– A)
e2A
2
=
e2A + 2 + e– 2A
e2A
= 1 + 2e– 2A + e– 4A = (1 + e– 2A)
vi)
(eA + e3B)(e–A + e–B)
e2B eA
2
1 + eA – B + e3B – A + e2B
=
e2BeA
= e–A – 2B + e– 3B + eB – 2A + e– A
4.3.5 Logarithms to general base
4.3.5A.
Find x if
i) 8 = log2 x
ii) 3 = log2 x
iv) 6 = log3 x
v) 4 = log3 x
Solution
-6-
iii) 4 = ln x
vi) 2 = ln x
All these questions are of the same type and rely on the inverse relation
If y = loga x then x = ay
i) If 8 = log2 x then x = 28 = 256
ii) If 3 = log2 x then x = 23 = 8
iii) If 4 = ln x = loge x then x = e4
iv) If 6 = log3 x then x = 36 = 729
v) If 4 = log3 x then x = 34 = 81
vi) If 2 = ln x = loge x then x = e2
4.3.5B.
Evaluate
i) ln e3
iv)
log9 81
vii)
ln e7
ii) log4 (256)
iii) log3 27
v) log4 2
vi) ln (e2)
2
viii) log3 (243)
Solution
Here we use the general result that
loga ax = x
So we must always express the number under the log as a power of the base of the
log.
i) ln e3 = loge e3 = 3
ii) log4 (256) = log4 44 = 4
iii) log3 27 = log3 33 = 3
iv) log9 81 = log9 92 = 2
1
v) log4 2 = log4 41/2 = 2
-7-
2
vi) ln (e2) = loge e4 = 4
vii) ln e7 = 7
viii) log3 (243) = log3 35 = 5
4.3.6 Manipulation of logarithms
4.3.6A.
Simplify as a single log
i) 2ln e4 + 3ln e3
iii)
loga x + loga(2y)
v)
2logax + 3ln x
ii) 3 log2 x + log2 x2
1
iv) ln(3x) – 2 ln (9x2)
vi) logax – log2ax
vii)
ln x + 2 logax2
Solution
In this question we use the results
ln ex = x ln e = x and loga xn = n loga x
i) 2ln e4 + 3ln e3 = 8 ln e + 9 ln e = 8 + 9 = 17
ii) 3 log2 x + log2 x2 = log2 x3 + log2 x2 = log2 (x3 x2) = log2 x5
iii) loga x + loga(2y) = loga(2xy)
1
1/2
iv) ln(3x) – 2 ln (9x2) = ln (3x) – ln (9x2)
= ln 3x – ln 3x = 0
v) In this question we must change both logs to the same base, which
might as well be e. We have
logb x
loga x = log a
b
So with b = e
ln x
loga x = ln a
-8-
Using this in the given expression gives
ln x
 2

2logax + 3ln x = 2 ln a + 3 ln x = ln a + 3 ln x = ln x(3 +2 /ln a)


(
)
logax
1 

logax – log2ax = logax – log 2a = logax 1 – log 2a
a
a 

vi)
 1– 1 


= loga x loga2a
ln x
ln x + 2 logax2 = ln x + 4 logax = ln x + 4 ln a
vii)
= ln
 1+ 4 
ln a

x

4.3.6B.
Expand each as a linear combination of numbers and logs in simplest form
i) ln (3x2y)
ii) log2 (8x2 y3)
iii) ln (eA/eB)
iv)
loga (ax ya)
v) log2a (8a3 x2 y4)
vi) ln (x2 y2 z2)
Solution
i) ln (3x2y) = ln 3 + ln (x2) + ln y = ln 3 + 2 ln x + ln y
ii) log2 (8x2 y3) = log2 8 + log2 (x2) + log2 (y3) = log2 8 + 2 log2 x + 3 log2 y
3 + 2 log2 x + 3 log2 y
iii)ln (eA/eB) = ln (eA) – ln (eB) = A – B
iv) loga (ax ya) = loga (ax) + loga (ya) = x + a loga y
v) log2a (8a3 x2 y4)= log2a (8a3) + log2a (x2) + log2a (y4)
= log2a (2a)3 + 2 log2a x + 4 log2a y = 3 + 2 log2a x + 4 log2a y
-9-
vi) ln (x2 y2 z2) = ln (x2) + ln (y2) + ln (z2) = 2 ln x + 2 ln y + 2 ln z
4.3.6C.
Simplify
2
aloga6 x2 ln (e3x)
2a x log2(4x)
i)
ii)
a(x–1) a2x a3
(ax)2 ax2 ln (e2a)
Solution
We need to take these slowly, step by step!
i) Breaking the expression up we have
aloga6 = 6
ln (e3x) = 3x
log2(4x) = log2(22x) = 2x
Now it is not so bad:
aloga6 x2 ln (e3x) 6 x2 (3x) 3 (3x)
= 2a x (2x) = 4a
2a x log2(4x)
9x
= 4a
ii) Again, break things up and take your time!
2
a(x–1) a2x a3
(ax)2 ax2 ln (e2a)
=
ax
2 – 2x + 1
a2x a3
2
a2x ax 2a
2
ax + 4
a– 2x+ 3
= 2x x2 = 2
2a a a
4.3.6D.
Evaluate
i)
log2 32
iv)
log5 625
vii)
log1/8 64
ii)
log10 100
iii) log7 49
v)
a1/2
vi) ln e2001
viii)
loga
ln 1
e
Solution
i) log2 32 = log2 25 = 5
- 10 -
ix) log8 2
ii) log10 100 = log10 102 = 2
iii) log7 49 = log7 72 = 2
iv) log5 625 = log5 54 = 4
1
v) loga a1/2 = 2
vi) ln e2001 = 2001
 1  –1
1 –2
vii) log1/8 64 = log1/8 64 = log1/8 8 = – 2
 
 
viii) ln 1 = ln e–1 = – 1
e
1
ix) log8 2 = log8 81/3 = 3
4.3.6E.
Simplify
i)
iv)
log 81
log 9
ii)
log 8
log 2
5 log 2– 3 log 32
v)
1
2 log 49
log 49
iii) log 343
Solution
log 81
log 92 2 log 9
i) log 9 = log 9 = log 9 = 2
log 8
3 log 2
ii) log 2 = log 2 = 3
log 49
log 72 2 log 7 2
iii) log 343 =
=
=
log 73 3 log 7 3
iv) 5 log 2– 3 log 32 = 5 log 2– 3 log 25 = 5 log 2 – 15 log 2 = – 10 log 2
- 11 -
1
v) 2 log 49 = log (491/2) = log 7
4.3.6F.
Given that
1
1
1
ln y = 2 ln (x + 1) – 2 ln (x – 1) + 3x + ln x + C
where C is an arbitrary constant, obtain an explicit expression for y in terms of x.
Solution
1
1
1
ln y = 2 ln (x + 1) – 2 ln (x – 1) + 3x + ln x + C
1
1
x + 1 1
= 2 ln  x – 1  + 2 ln x2 + 2 ln e2(3x + C)



1
x + 1

= 2 ln  x – 1  x2 e2(3x + C) = ln x




x + 1 (3x + C)

x–1 e

So
1
y =x
x + 1 (3x + C)
x–1 e
and hence
1
y=x
x – 1 – (3x + C)
x+1 e
4.3.7 Some applications of logarithms
4.3.7A.
Solve the following equations, giving your answers to 3 decimal places.
i) 3x = 16
ii) 42x = 9
iii) 4 5– 2x = 3 7x – 2
Solution
i) If 3x = 16 then taking logs base e we have
ln (3x) = x ln 3 = ln 16
so
- 12 -
iv) 3x = 42x – 1
ln 16
x = ln 3 = 2.528 to 3dp
ii) ln(42x) = 2x ln 4 = ln 9. So
ln 9
x = 2 ln 4 = 0.792 to 3dp
iii) Taking natural logs of 4 5– 2x = 3 7x – 2 gives
ln 4 – 2x ln 5 = ln 3 + (x – 2) ln 7
Gathering like terms together gives
4 49
x(2 ln 5 + ln 7) = ln 4 + 2 ln 7 – ln 3 = ln  3 


So
4 49
ln  3 


x = 2 ln 5 + ln 7 = 0.809 to 3dp
iv) From 3x = 42x – 1 we get
x ln 3 = (2x – 1) ln 4
or
(2 ln 4 – ln 3)x = ln 4
giving
ln 4
x = 2 ln 4 – ln 3 = 0.828 to 3dp
4.3.7B.
Convert the following equations to straight line form
i) y = 4 x7
ii) y = 3 x– 4 iii) y =
5
x3
iv) y = 20 e– 2x v) y = 2 4x – 1
Solution
i) If y = 4 x7 then taking logs gives
ln y = ln 4 + 7 ln x
So if we put X = ln x and Y = ln y then we get the linear form
Y = 7X + ln 4
- 13 -
ii) Taking logs of y = 3 x– 4 gives
ln y = ln 3 – 4 ln x
or with Y = ln y and X = ln x
Y = ln 3 – 4X
iii) y =
5
= 5x– 3 and by the same process as in ii) we will get
x3
Y = ln 5 – 3X
iv) In the case of y = 20 e– 2x taking natural logs gives
ln y = ln 20 – 2x
So with X = x and Y = ln y we get
Y = ln 20 – 2X
v) y = 2 4x – 1 gives, on taking logs
ln y = ln(2 4x – 1) = (4x – 1) ln 2 = (4 ln 2) x – ln 2
So with X = x and Y = ln y we get
Y = (4 ln 2) X – ln 2
- 14 -