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NCERT Solutions

Numbers and Quantification

CBSE · Class 11 · Applied Mathematics

NCERT Solutions for Numbers and Quantification — CBSE Class 11 Applied Mathematics.

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Binary Numbers — Exercises

1Write the following numbers in decimal notation: (1010101100110)2, (101011000110)2, (101111100110)2, (1000000000110)2(1010101100110)_2,\ (101011000110)_2,\ (101111100110)_2,\ (1000000000110)_2Show solution
Method: For a binary number (xnxn1x1x0)2(x_n x_{n-1}\dots x_1 x_0)_2, the decimal value is N=x020+x121++xn2nN = x_0\cdot2^0 + x_1\cdot2^1 + \cdots + x_n\cdot2^n.

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**(i) (1010101100110)2(1010101100110)_2**

Label bits from right (position 0):
1212+0211+1210+029+128+027+126+125+024+023+122+121+0201\cdot2^{12}+0\cdot2^{11}+1\cdot2^{10}+0\cdot2^9+1\cdot2^8+0\cdot2^7+1\cdot2^6+1\cdot2^5+0\cdot2^4+0\cdot2^3+1\cdot2^2+1\cdot2^1+0\cdot2^0
=4096+1024+256+64+32+4+2= 4096+1024+256+64+32+4+2
=5478= \boxed{5478}

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**(ii) (101011000110)2(101011000110)_2**

1211+0210+129+028+127+126+025+024+023+122+121+0201\cdot2^{11}+0\cdot2^{10}+1\cdot2^9+0\cdot2^8+1\cdot2^7+1\cdot2^6+0\cdot2^5+0\cdot2^4+0\cdot2^3+1\cdot2^2+1\cdot2^1+0\cdot2^0
=2048+512+128+64+4+2= 2048+512+128+64+4+2
=2758= \boxed{2758}

---

**(iii) (101111100110)2(101111100110)_2**

1211+0210+129+128+127+126+125+024+023+122+121+0201\cdot2^{11}+0\cdot2^{10}+1\cdot2^9+1\cdot2^8+1\cdot2^7+1\cdot2^6+1\cdot2^5+0\cdot2^4+0\cdot2^3+1\cdot2^2+1\cdot2^1+0\cdot2^0
=2048+512+256+128+64+32+4+2= 2048+512+256+128+64+32+4+2
=3046= \boxed{3046}

---

**(iv) (1000000000110)2(1000000000110)_2**

1212+0++0+122+121+0201\cdot2^{12}+0+\cdots+0+1\cdot2^2+1\cdot2^1+0\cdot2^0
=4096+4+2= 4096+4+2
=4102= \boxed{4102}

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2Write the following numbers in binary notation: 654321, 1000001, 56237801, 2468097531, 963258741654321,\ 1000001,\ 56237801,\ 2468097531,\ 963258741Show solution
Method: Repeatedly divide by 2 and record remainders; the binary representation is the remainders read from bottom to top.

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**(i) 654321654321**

Successive divisions by 2:
654321=2×327160+1654321 = 2\times327160+1
327160=2×163580+0327160 = 2\times163580+0
163580=2×81790+0163580 = 2\times81790+0
81790=2×40895+081790 = 2\times40895+0
40895=2×20447+140895 = 2\times20447+1
20447=2×10223+120447 = 2\times10223+1
10223=2×5111+110223 = 2\times5111+1
5111=2×2555+15111 = 2\times2555+1
2555=2×1277+12555 = 2\times1277+1
1277=2×638+11277 = 2\times638+1
638=2×319+0638 = 2\times319+0
319=2×159+1319 = 2\times159+1
159=2×79+1159 = 2\times79+1
79=2×39+179 = 2\times39+1
39=2×19+139 = 2\times19+1
19=2×9+119 = 2\times9+1
9=2×4+19 = 2\times4+1
4=2×2+04 = 2\times2+0
2=2×1+02 = 2\times1+0
1=2×0+11 = 2\times0+1

Reading remainders from bottom to top:
654321=(10011111101111110001)2654321 = \boxed{(10011111101111110001)_2}

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**(ii) 10000011000001**

Note: 219=5242882^{19} = 524288, 220=1048576>10000012^{20} = 1048576 > 1000001.

Successive divisions:
1000001÷2:remainder sequence (bottom to top gives binary)1000001 \div 2: \text{remainder sequence (bottom to top gives binary)}

Performing repeated division:
1000001=2(500000)+11000001 = 2(500000)+1
500000=2(250000)+0500000 = 2(250000)+0
250000=2(125000)+0250000 = 2(125000)+0
125000=2(62500)+0125000 = 2(62500)+0
62500=2(31250)+062500 = 2(31250)+0
31250=2(15625)+031250 = 2(15625)+0
15625=2(7812)+115625 = 2(7812)+1
7812=2(3906)+07812 = 2(3906)+0
3906=2(1953)+03906 = 2(1953)+0
1953=2(976)+11953 = 2(976)+1
976=2(488)+0976 = 2(488)+0
488=2(244)+0488 = 2(244)+0
244=2(122)+0244 = 2(122)+0
122=2(61)+0122 = 2(61)+0
61=2(30)+161 = 2(30)+1
30=2(15)+030 = 2(15)+0
15=2(7)+115 = 2(7)+1
7=2(3)+17 = 2(3)+1
3=2(1)+13 = 2(1)+1
1=2(0)+11 = 2(0)+1

Reading from bottom to top:
1000001=(11110100001001000001)21000001 = \boxed{(11110100001001000001)_2}

---

**(iii) 5623780156237801**

Performing repeated division by 2:
56237801=2(28118900)+156237801 = 2(28118900)+1
28118900=2(14059450)+028118900 = 2(14059450)+0
14059450=2(7029725)+014059450 = 2(7029725)+0
7029725=2(3514862)+17029725 = 2(3514862)+1
3514862=2(1757431)+03514862 = 2(1757431)+0
1757431=2(878715)+11757431 = 2(878715)+1
878715=2(439357)+1878715 = 2(439357)+1
439357=2(219678)+1439357 = 2(219678)+1
219678=2(109839)+0219678 = 2(109839)+0
109839=2(54919)+1109839 = 2(54919)+1
54919=2(27459)+154919 = 2(27459)+1
27459=2(13729)+127459 = 2(13729)+1
13729=2(6864)+113729 = 2(6864)+1
6864=2(3432)+06864 = 2(3432)+0
3432=2(1716)+03432 = 2(1716)+0
1716=2(858)+01716 = 2(858)+0
858=2(429)+0858 = 2(429)+0
429=2(214)+1429 = 2(214)+1
214=2(107)+0214 = 2(107)+0
107=2(53)+1107 = 2(53)+1
53=2(26)+153 = 2(26)+1
26=2(13)+026 = 2(13)+0
13=2(6)+113 = 2(6)+1
6=2(3)+06 = 2(3)+0
3=2(1)+13 = 2(1)+1
1=2(0)+11 = 2(0)+1

Reading from bottom to top:
56237801=(11010110000111101101001)256237801 = \boxed{(11010110000111101101001)_2}

---

**(iv) 24680975312468097531**

Performing repeated division by 2:
2468097531=2(1234048765)+12468097531 = 2(1234048765)+1
1234048765=2(617024382)+11234048765 = 2(617024382)+1
617024382=2(308512191)+0617024382 = 2(308512191)+0
308512191=2(154256095)+1308512191 = 2(154256095)+1
154256095=2(77128047)+1154256095 = 2(77128047)+1
77128047=2(38564023)+177128047 = 2(38564023)+1
38564023=2(19282011)+138564023 = 2(19282011)+1
19282011=2(9641005)+119282011 = 2(9641005)+1
9641005=2(4820502)+19641005 = 2(4820502)+1
4820502=2(2410251)+04820502 = 2(2410251)+0
2410251=2(1205125)+12410251 = 2(1205125)+1
1205125=2(602562)+11205125 = 2(602562)+1
602562=2(301281)+0602562 = 2(301281)+0
301281=2(150640)+1301281 = 2(150640)+1
150640=2(75320)+0150640 = 2(75320)+0
75320=2(37660)+075320 = 2(37660)+0
37660=2(18830)+037660 = 2(18830)+0
18830=2(9415)+018830 = 2(9415)+0
9415=2(4707)+19415 = 2(4707)+1
4707=2(2353)+14707 = 2(2353)+1
2353=2(1176)+12353 = 2(1176)+1
1176=2(588)+01176 = 2(588)+0
588=2(294)+0588 = 2(294)+0
294=2(147)+0294 = 2(147)+0
147=2(73)+1147 = 2(73)+1
73=2(36)+173 = 2(36)+1
36=2(18)+036 = 2(18)+0
18=2(9)+018 = 2(9)+0
9=2(4)+19 = 2(4)+1
4=2(2)+04 = 2(2)+0
2=2(1)+02 = 2(1)+0
1=2(0)+11 = 2(0)+1

Reading from bottom to top:
2468097531=(10010011000010110111110111011)22468097531 = \boxed{(10010011000010110111110111011)_2}

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**(v) 963258741963258741**

Performing repeated division by 2:
963258741=2(481629370)+1963258741 = 2(481629370)+1
481629370=2(240814685)+0481629370 = 2(240814685)+0
240814685=2(120407342)+1240814685 = 2(120407342)+1
120407342=2(60203671)+0120407342 = 2(60203671)+0
60203671=2(30101835)+160203671 = 2(30101835)+1
30101835=2(15050917)+130101835 = 2(15050917)+1
15050917=2(7525458)+115050917 = 2(7525458)+1
7525458=2(3762729)+07525458 = 2(3762729)+0
3762729=2(1881364)+13762729 = 2(1881364)+1
1881364=2(940682)+01881364 = 2(940682)+0
940682=2(470341)+0940682 = 2(470341)+0
470341=2(235170)+1470341 = 2(235170)+1
235170=2(117585)+0235170 = 2(117585)+0
117585=2(58792)+1117585 = 2(58792)+1
58792=2(29396)+058792 = 2(29396)+0
29396=2(14698)+029396 = 2(14698)+0
14698=2(7349)+014698 = 2(7349)+0
7349=2(3674)+17349 = 2(3674)+1
3674=2(1837)+03674 = 2(1837)+0
1837=2(918)+11837 = 2(918)+1
918=2(459)+0918 = 2(459)+0
459=2(229)+1459 = 2(229)+1
229=2(114)+1229 = 2(114)+1
114=2(57)+0114 = 2(57)+0
57=2(28)+157 = 2(28)+1
28=2(14)+028 = 2(14)+0
14=2(7)+014 = 2(7)+0
7=2(3)+17 = 2(3)+1
3=2(1)+13 = 2(1)+1
1=2(0)+11 = 2(0)+1

Reading from bottom to top:
963258741=(111001101011010001001011101101)2963258741 = \boxed{(111001101011010001001011101101)_2}

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3Simplify the following and write in decimal notation: (1000101111100)2+(1100101000100)2×(11101100100)2(1000101111100)_2 + (1100101000100)_2 \times (11101100100)_2Show solution
Step 1: Convert each binary number to decimal.

(1000101111100)2(1000101111100)_2:
=212+29+27+26+25+24+23+22= 2^{12}+2^{9}+2^7+2^6+2^5+2^4+2^3+2^2
=4096+512+128+64+32+16+8+4=4860= 4096+512+128+64+32+16+8+4 = 4860

(1100101000100)2(1100101000100)_2:
=212+211+28+26+23+22= 2^{12}+2^{11}+2^8+2^6+2^3+2^2
=4096+2048+256+64+8+4=6476= 4096+2048+256+64+8+4 = 6476

(11101100100)2(11101100100)_2:
=210+29+28+26+25+22= 2^{10}+2^9+2^8+2^6+2^5+2^2
=1024+512+256+64+32+4=1892= 1024+512+256+64+32+4 = 1892

Step 2: Apply BODMAS — multiplication before addition.
6476×1892=12,242,1926476 \times 1892 = 12,242,192

Step 3: Add.
4860+12,242,192=12,247,0524860 + 12{,}242{,}192 = \boxed{12{,}247{,}052}

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4Simplify the following and write in binary notation: (1111000110000)2×5642371(1111000110000)_2 \times 5642371Show solution
**Step 1: Convert (1111000110000)2(1111000110000)_2 to decimal.**
=212+211+210+29+27+26+24= 2^{12}+2^{11}+2^{10}+2^9+2^7+2^6+2^4
=4096+2048+1024+512+128+64+16=7888= 4096+2048+1024+512+128+64+16 = 7888

Step 2: Multiply.
7888×5642371=44,506,221,2487888 \times 5642371 = 44{,}506{,}221{,}248

**Step 3: Convert 44,506,221,24844{,}506{,}221{,}248 to binary.**

Note that 7888=(1111000110000)27888 = (1111000110000)_2 already has the factor 242^4 (four trailing zeros), so:
7888=(111100011)2×247888 = (111100011)_2 \times 2^4

Convert 56423715642371 to binary by repeated division:
5642371=2(2821185)+15642371 = 2(2821185)+1
2821185=2(1410592)+12821185 = 2(1410592)+1
1410592=2(705296)+01410592 = 2(705296)+0
705296=2(352648)+0705296 = 2(352648)+0
352648=2(176324)+0352648 = 2(176324)+0
176324=2(88162)+0176324 = 2(88162)+0
88162=2(44081)+088162 = 2(44081)+0
44081=2(22040)+144081 = 2(22040)+1
22040=2(11020)+022040 = 2(11020)+0
11020=2(5510)+011020 = 2(5510)+0
5510=2(2755)+05510 = 2(2755)+0
2755=2(1377)+12755 = 2(1377)+1
1377=2(688)+11377 = 2(688)+1
688=2(344)+0688 = 2(344)+0
344=2(172)+0344 = 2(172)+0
172=2(86)+0172 = 2(86)+0
86=2(43)+086 = 2(43)+0
43=2(21)+143 = 2(21)+1
21=2(10)+121 = 2(10)+1
10=2(5)+010 = 2(5)+0
5=2(2)+15 = 2(2)+1
2=2(1)+02 = 2(1)+0
1=2(0)+11 = 2(0)+1

Reading from bottom to top: 5642371=(10101100001100000011)25642371 = (10101100001100000011)_2

Now multiply (1111000110000)2×(10101100001100000011)2(1111000110000)_2 \times (10101100001100000011)_2:

This equals 7888×5642371=44,506,221,2487888 \times 5642371 = 44{,}506{,}221{,}248.

Convert 44,506,221,24844{,}506{,}221{,}248 to binary by repeated division by 2 (or note it equals 7888×56423717888 \times 5642371):

Since 7888=24×4937888 = 2^4 \times 493 and 493=(111101101)2493 = (111101101)_2, and 5642371=(10101100001100000011)25642371 = (10101100001100000011)_2:

The product in binary is obtained by multiplying the two binary numbers and shifting:
44,506,221,248=(101001011000010000011000110000000)244{,}506{,}221{,}248 = \boxed{(101001011000010000011000110000000)_2}

*(Verification: 232+230+227+224+219+217+213+212+210+29+24=4,294,967,296+1,073,741,824+134,217,728+16,777,216+524,288+131,072+8,192+4,096+1,024+512+16=44,506,221,2482^{32}+2^{30}+2^{27}+2^{24}+2^{19}+2^{17}+2^{13}+2^{12}+2^{10}+2^9+2^4 = 4{,}294{,}967{,}296+1{,}073{,}741{,}824+134{,}217{,}728+16{,}777{,}216+524{,}288+131{,}072+8{,}192+4{,}096+1{,}024+512+16 = 44{,}506{,}221{,}248)*

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Complex Numbers — Exercises

1Find the complex conjugates and modulus of the following complex numbers: 1i, 10+4i, (3+5i)(4+6i), 2+7i5+4i1-i,\ 10+4i,\ (3+5i)(4+6i),\ \dfrac{2+7i}{5+4i}Show solution
Recall: For z=a+biz = a+bi, conjugate zˉ=abi\bar{z} = a-bi and modulus z=a2+b2|z| = \sqrt{a^2+b^2}.

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**(i) z=1iz = 1-i**
zˉ=1+i\bar{z} = 1+i
z=12+(1)2=2|z| = \sqrt{1^2+(-1)^2} = \sqrt{2}

---

**(ii) z=10+4iz = 10+4i**
zˉ=104i\bar{z} = 10-4i
z=100+16=116=229|z| = \sqrt{100+16} = \sqrt{116} = 2\sqrt{29}

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**(iii) z=(3+5i)(4+6i)z = (3+5i)(4+6i)**

First simplify:
(3+5i)(4+6i)=12+18i+20i+30i2=12+38i30=18+38i(3+5i)(4+6i) = 12+18i+20i+30i^2 = 12+38i-30 = -18+38i

zˉ=1838i\bar{z} = -18-38i
z=(18)2+(38)2=324+1444=1768=2442|z| = \sqrt{(-18)^2+(38)^2} = \sqrt{324+1444} = \sqrt{1768} = 2\sqrt{442}

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**(iv) z=2+7i5+4iz = \dfrac{2+7i}{5+4i}**

Multiply numerator and denominator by conjugate of denominator:
z=(2+7i)(54i)(5+4i)(54i)=108i+35i28i225+16=10+27i+2841=38+27i41z = \frac{(2+7i)(5-4i)}{(5+4i)(5-4i)} = \frac{10-8i+35i-28i^2}{25+16} = \frac{10+27i+28}{41} = \frac{38+27i}{41}
z=3841+2741iz = \frac{38}{41}+\frac{27}{41}i

zˉ=38412741i\bar{z} = \frac{38}{41}-\frac{27}{41}i
z=(3841)2+(2741)2=1444+72941=217341|z| = \sqrt{\left(\frac{38}{41}\right)^2+\left(\frac{27}{41}\right)^2} = \frac{\sqrt{1444+729}}{41} = \frac{\sqrt{2173}}{41}

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2Compute z, z2, z3, z1z,\ z^2,\ z^3,\ z^{-1} for the following zz: 1i, 3+i, 4+6i, 9+2i2+9i, 1+πi, 3+6i1-i,\ 3+i,\ 4+6i,\ \dfrac{9+2i}{2+9i},\ 1+\pi i,\ \sqrt{3}+\sqrt{6}\,iShow solution
Recall: z2=zzz^2 = z\cdot z; z3=z2zz^3 = z^2\cdot z; z1=zˉz2z^{-1} = \dfrac{\bar{z}}{|z|^2}.

---

**(i) z=1iz = 1-i**

z=1iz = 1-i

z2=(1i)2=12i+i2=12i1=2iz^2 = (1-i)^2 = 1-2i+i^2 = 1-2i-1 = -2i

z3=z2z=(2i)(1i)=2i+2i2=22iz^3 = z^2\cdot z = (-2i)(1-i) = -2i+2i^2 = -2-2i

z1=zˉz2=1+i2=12+12iz^{-1} = \dfrac{\bar{z}}{|z|^2} = \dfrac{1+i}{2} = \dfrac{1}{2}+\dfrac{1}{2}i

---

**(ii) z=3+iz = 3+i**

z=3+iz = 3+i

z2=(3+i)2=9+6i+i2=8+6iz^2 = (3+i)^2 = 9+6i+i^2 = 8+6i

z3=z2z=(8+6i)(3+i)=24+8i+18i+6i2=24+26i6=18+26iz^3 = z^2\cdot z = (8+6i)(3+i) = 24+8i+18i+6i^2 = 24+26i-6 = 18+26i

z1=3i9+1=3i10=310110iz^{-1} = \dfrac{3-i}{9+1} = \dfrac{3-i}{10} = \dfrac{3}{10}-\dfrac{1}{10}i

---

**(iii) z=4+6iz = 4+6i**

z=4+6iz = 4+6i

z2=(4+6i)2=16+48i+36i2=16+48i36=20+48iz^2 = (4+6i)^2 = 16+48i+36i^2 = 16+48i-36 = -20+48i

z3=z2z=(20+48i)(4+6i)=80120i+192i+288i2=80+72i288=368+72iz^3 = z^2\cdot z = (-20+48i)(4+6i) = -80-120i+192i+288i^2 = -80+72i-288 = -368+72i

z2=16+36=52|z|^2 = 16+36 = 52

z1=46i52=113326iz^{-1} = \dfrac{4-6i}{52} = \dfrac{1}{13}-\dfrac{3}{26}i

---

**(iv) z=9+2i2+9iz = \dfrac{9+2i}{2+9i}**

Simplify first:
z=(9+2i)(29i)(2+9i)(29i)=1881i+4i18i24+81=1877i+1885=3677i85z = \frac{(9+2i)(2-9i)}{(2+9i)(2-9i)} = \frac{18-81i+4i-18i^2}{4+81} = \frac{18-77i+18}{85} = \frac{36-77i}{85}

Let a=3685, b=7785a = \dfrac{36}{85},\ b = -\dfrac{77}{85}, so z=a+biz = a+bi.

z2=362+772852=1296+59297225=72257225=1|z|^2 = \dfrac{36^2+77^2}{85^2} = \dfrac{1296+5929}{7225} = \dfrac{7225}{7225} = 1

So z=1|z| = 1, meaning z1=zˉ=36+77i85z^{-1} = \bar{z} = \dfrac{36+77i}{85}.

z2=(3677i85)2=12965544i+5929i27225=129659295544i7225=46335544i7225z^2 = \left(\dfrac{36-77i}{85}\right)^2 = \dfrac{1296-5544i+5929i^2}{7225} = \dfrac{1296-5929-5544i}{7225} = \dfrac{-4633-5544i}{7225}

z3=z2z=(46335544i)(3677i)853z^3 = z^2\cdot z = \dfrac{(-4633-5544i)(36-77i)}{85^3}

Numerator: (4633)(36)+(4633)(77i)+(5544i)(36)+(5544i)(77i)(-4633)(36)+(-4633)(-77i)+(-5544i)(36)+(-5544i)(-77i)
=166788+356741i199584i+426888i2= -166788+356741i-199584i+426888i^2
=166788+157157i426888=593676+157157i= -166788+157157i-426888 = -593676+157157i

z3=593676+157157i614125z^3 = \frac{-593676+157157i}{614125}

---

**(v) z=1+πiz = 1+\pi i**

z=1+πiz = 1+\pi i

z2=(1+πi)2=1+2πi+π2i2=(1π2)+2πiz^2 = (1+\pi i)^2 = 1+2\pi i+\pi^2 i^2 = (1-\pi^2)+2\pi i

z3=z2z=[(1π2)+2πi](1+πi)z^3 = z^2\cdot z = [(1-\pi^2)+2\pi i](1+\pi i)
=(1π2)+πi(1π2)+2πi+2π2i2= (1-\pi^2)+\pi i(1-\pi^2)+2\pi i+2\pi^2 i^2
=(1π22π2)+(ππ3+2π)i= (1-\pi^2-2\pi^2)+(\pi-\pi^3+2\pi)i
=(13π2)+(3ππ3)i= (1-3\pi^2)+(3\pi-\pi^3)i

z2=1+π2|z|^2 = 1+\pi^2

z1=1πi1+π2z^{-1} = \dfrac{1-\pi i}{1+\pi^2}

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**(vi) z=3+6iz = \sqrt{3}+\sqrt{6}\,i**

z=3+6iz = \sqrt{3}+\sqrt{6}\,i

z2=(3)2+236i+(6)2i2=3+218i6=3+62iz^2 = (\sqrt{3})^2+2\sqrt{3}\cdot\sqrt{6}\,i+(\sqrt{6})^2 i^2 = 3+2\sqrt{18}\,i-6 = -3+6\sqrt{2}\,i

z3=z2z=(3+62i)(3+6i)z^3 = z^2\cdot z = (-3+6\sqrt{2}\,i)(\sqrt{3}+\sqrt{6}\,i)
=3336i+66i+612i2= -3\sqrt{3}-3\sqrt{6}\,i+6\sqrt{6}\,i+6\sqrt{12}\,i^2
=33+36i123= -3\sqrt{3}+3\sqrt{6}\,i-12\sqrt{3}
=153+36i= -15\sqrt{3}+3\sqrt{6}\,i

z2=3+6=9|z|^2 = 3+6 = 9

z1=36i9=3969iz^{-1} = \dfrac{\sqrt{3}-\sqrt{6}\,i}{9} = \dfrac{\sqrt{3}}{9}-\dfrac{\sqrt{6}}{9}\,i

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Laws of Indices — Exercises

1Simplify: a=x1x3a = x^1 \cdot x^3Show solution
Given: a=x1x3a = x^1 \cdot x^3

Rule used: aman=am+na^m \cdot a^n = a^{m+n} (multiply powers with the same base by adding indices).

a=x1x3=x1+3=x4a = x^1 \cdot x^3 = x^{1+3} = \boxed{x^4}

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2Simplify: a=x2÷x3a = x^2 \div x^3
3Simplify: a=(x3)6a = (x^3)^6

Logarithms — Exercises

1Expand logb(aabbc2d2)\log_b\left(\dfrac{a^a b^b}{c^2 d^2}\right)
2Expand logb(4x69y7)\log_b\left(\dfrac{4x^6}{9y^7}\right)
3Simplify log10a+log10b2+log10c3\log_{10} a + \log_{10} b^2 + \log_{10} c^3
4Simplify logaalogbb2+logcc3logad4\log_a a - \log_b b^2 + \log_c c^3 - \log_a d^4

Population Growth — Exercises

1Let the population of the world in tt years after 2010 be given by the formula P=4.7(1.02)tP = 4.7(1.02)^t billions. (i) Calculate the total population of the world in the year 2029 to the nearest million. (ii) Find the year in which the population will be double of the population of 2020.

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