Chapter 4 Solutions to Reinforcement Exercises in Exponentials and Logarithms

CHAPTER 4 SOLUTIONS TO REINFORCEMENT EXERCISES IN EXPONENTIALS AND LOGARITHMS

4.3.1y = 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 = xlies sandwiched between those of y = 3x and y = 4x as shown.

4.3.2The general exponential function ax

Simplify

i)ii)

iii)iv)

v)vi)

vii)

Solution

i) It is best to first express everything in terms of powers of 2, then:

= = = = 1

ii) In this case we can express everything in terms of powers of 2 and 3:

=

Now collect terms:

= 2x/222(x + 1)2-5x/23x/23x + 13–3x/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)= = = 1

iv) In this case, multiply top and bottom by (x2 + 1)to obtain

=

= = –

v)=

= ex e–x2e– x+ 1 e– x2 –2x–1 = e– 2x2e– 2x

= e– 2x(x + 1)

vi)=

= a3 a– x2 – 2x - 1ax2 a2x = a3– x2 – 2x – 1+ x2 + 2x = a2

vii) Remember the trig identities cos 2x = 2 cos2x – 1 and cos2x + sin2x = 1:

=

= a2cos2x – 1 a– cos2xa– 3 + sin2x = a2cos2x – 1 a– cos2xa– 3 + sin2x

= a2cos2x – 1– cos2x – 3 + sin2x = a – 3

4.3.3The 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

£ C525600

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

£ C31536000

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

ex = 1 + x + + + … + + …

i) For x = 1 we therefore have

e = 1 + 1 + + + + … + + …

= 1 + 1 + + + + ...

= 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

e0.1 = e = 1 + 0.1 + + + + … + + …

= 1 + 0.1 + 0.005 + 0.00017 + 0.000004 ...

= 1.1052 to 4 dp

iii) For x = 2 we have

e2 = e = 1 + 2 + + + + … + + …

= 1 + 2 + ...

= 7.3891 to 4 dp

4.3.3C.

Sketch the curves

i)y = ex – 1ii)y = 1 – e–x

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

4.3.4Manipulation of the exponential function.

Simplify

i)eA (eB)2 e3Cii)iii)

iv)v)vi)

Solution

i)eA (eB)2 e3C = eA e2B e3C = eA + 2B + 3C

ii)= ex e3ye2xex e– y = ex e2xexe3ye– y = e4x e2y = e2(2x + y)

iii) = = e3x ex2 e2ye– 4y e– x– 3 = e3x + x2 + 2y – 4y - x - 3

= ex2 + 2x – 2y - 3

iv)= = e2B eA eB–2eB – 2 e– 2A

= e– A + 4B – 4

v)= =

= 1 + 2e– 2A + e– 4A = (1 + e– 2A)2

vi) =

= e–A – 2B + e– 3B + eB – 2A + e– A

4.3.5Logarithms to general base

4.3.5A.

Find x if

i)8 = log2 xii)3 = log2 xiii)4 = ln x

iv)6 = log3 xv)4 = log3 xvi)2 = ln x

Solution

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 e3ii)log4 (256)iii)log3 27

iv)log9 81v)log4 2vi)ln (e2)2

vii)ln e7viii)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

v) log4 2 = log4 41/2 =

vi) ln (e2)2 = loge e4 = 4

vii) ln e7 = 7

viii) log3 (243) = log3 35 = 5

4.3.6Manipulation of logarithms

4.3.6A.

Simplify as a single log

i)2ln e4 + 3ln e3ii)3 log2 x + log2 x2

iii)loga x + loga(2y)iv)ln(3x) – ln (9x2)

v)2logax + 3ln xvi)logax – log2axvii)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= log2 x5

iii) loga x + loga(2y) = loga(2xy)

iv) ln(3x) – ln (9x2) = ln (3x) – ln 1/2 = 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

loga x =

So with b = e

loga x =

Using this in the given expression gives

2logax + 3ln x = 2 + 3 ln x = ln x = ln

vi)logax – log2ax = logax – = logax

= loga

vii)ln x + 2 logax2 = ln x + 4 logax = ln x + 4 = ln

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

vi) ln (x2 y2 z2) =ln (x2) +ln (y2) +ln (z2) = 2 ln x + 2 ln y + 2 ln z

4.3.6C.

Simplify

i)ii)

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:

= =

=

ii) Again, break things up and take your time!

=

= =

4.3.6D.

Evaluate

i)log2 32ii)log10 100iii)log7 49

iv)log5 625v)loga avi)ln e2001

vii)log64viii)ln ix)log8 2

Solution

i) log2 32 = log2 25 = 5

ii) log10 100 = log10 102 = 2

iii) log7 49 = log7 72 = 2

iv) log5 625 = log5 54 = 4

v) loga a=

vi) ln e2001 = 2001

vii) log64 = log–1 = log–2 = – 2

viii) ln = ln e–1 = – 1

ix) log8 2 = log8 8=

4.3.6E.

Simplify

i)ii)iii)

iv)5 log 2– 3 log 32v)log 49

Solution

i) = = = 2

ii) = = 3

iii) = = =

iv) 5 log 2– 3 log 32 = 5 log 2– 3 log 25 = 5 log 2 – 15 log 2 = – 10 log 2

v) log 49 = log = log 7

4.3.6F.

Given that

ln = ln (x + 1) – ln (x – 1) + 3x +ln x + C

where C is an arbitrary constant, obtain an explicit expression for y in terms of x.

Solution

ln = ln (x + 1) – ln (x – 1) + 3x +ln x + C

= ln + ln x2 + ln e2(3x + C)

= ln = ln

So

= xe(3x + C)

and hence

y =e– (3x + C)

4.3.7Some 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 iv) 3x = 42x – 1

Solution

i) If 3x = 16 then taking logs base e we have

ln = x ln 3 = ln 16

so

x = = 2.528 to 3dp

ii) ln= 2x ln 4 = ln 9. So

x = = 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

x(2 ln 5 + ln 7) = ln 4 + 2 ln 7 – ln 3 = ln

So

x == 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

x = = 0.828 to 3dp

4.3.7B.

Convert the following equations to straight line form

i) y = 4 x7ii) y = 3 x– 4iii) y = 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

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 = = 5x– 3 and by the same process as in ii) we will get

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= (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

- 1 -