정답과 풀이

(1)

수학

문제기본서

정답과 풀이

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(2)

10

00 답 ㄱ, ㄴ

11

00 {Ü"Ã(-2)Ý`}Ü``=(Ü"Å2Ý`)Ü`=2Ý`=16

답 16

12

00 (¡'16)Û`=¡"16Û`=¡"(2Ý`)Û`=¡"2¡`=2

답 2

13

00 Ü'4_Ü'16=Ü'Ä4_16=Ü'64=Ü"Å4Ü`=4

답 4

14

00 ^Ý^'80

Ý'5 =Ý®É:¥5¼:=Ý'16=Ý"Å2Ý`=2

답 2

15

00 Ü"Ã'¶729_"Ã'¶256‌‌=ß'¶729_Ý'¶256=ß"Å3ß`_Ý"Å4Ý`‌ ‌

=3_4=12

답 12

16

00 답 1 00

17

답 1

18

00 답;1Á6; 00

19

답;2Á5;

20

00 답 9 00

21

답 81

22

00 답;4!; 00

23

답;5$;

24

00 답 -;3@; 00

25

답;3!;

26

00 (a^;4#;)Û`_a^;4!;=a^;2#;_a^;4!;=a^;2#;+;4!;=a^;4&;

답 a^;4&;

27

00 (aÜ`bÛ`)^;1Á2;_(a¹³b^;4!;)Ý`‌‌=a^;4!;b^;6!;_a^;3$;b‌ ‌

=a¹^{4 +;3$;}b^;6!;+1=a^;1!2(;b^;6&;

답 a¹⁹¹²b^;6&;

28

00 ("ÅaÜ`_Þ'a_a^-;2!;)^;3!;‌‌=(a^;2#;_a^;5!;_a^-;2!;)¹³‌‌

=(a;2#;+;5!;-;2!;)¹³=(a^{;5^;})^;3!;=a^;5@;

답 a^;5@;

29

00 (a^-;4#;)Û`_'aÖa^;4#;=a^-;2#;_a^;2!;Öa^;4#;=a-;2#;+;2!;-;4#;=a^-;4&;

답 a^-;4&;

01 지수

^Ⅰ.^{지수함수와 로그함수}

본문 7쪽

교과서 문제

^정^/^복^/^하^/^기

1

000 -8의‌세제곱근을‌x라‌하면‌xÜ`=-8이므로 xÜ`+8=0,‌(x+2)(xÛ`-2x+4)=0

∴‌x=-2‌또는‌x=1Ñ'3‌i

따라서‌-8의‌세제곱근‌중‌실수인‌것은‌-2이다.

답 -2

2

000 81의‌네제곱근을‌x라‌하면‌xÝ`=81이므로 xÝ`-81=0,‌(xÛ`+9)(xÛ`-9)=0

(xÛ`+9)(x+3)(x-3)=0

∴‌x=Ñ3i‌또는‌x=Ñ3

따라서‌81의‌네제곱근‌중‌실수인‌것은‌-3,‌3이다.

답 -3, 3

3

000 ‌‌0.027의‌세제곱근을‌x라‌하면‌xÜ`=0.027이므로 xÜ`-0.027=0,‌(x-0.3)(xÛ`+0.3x+0.09)=0 이때‌xÛ`+0.3x+0.09=0은‌실근을‌갖지‌않으므로 0.027의‌세제곱근‌중‌실수인‌것은‌0.3이다.

답 0.3

4

000 ‌‌(-2)Ý`=16의‌네제곱근을‌x라‌하면‌xÝ`=16이므로 xÝ`-16=0,‌(xÛ`+4)(xÛ`-4)=0

(xÛ`+4)(x+2)(x-2)=0

따라서‌(-2)Ý`의‌네제곱근‌중‌실수인‌것은‌-2,‌2이다.

답 -2, 2

5

000 -16의‌네제곱근을‌x라‌하면‌xÝ`=-16이므로‌x의‌값‌

중‌실수인‌것은‌없다.

답없다.

6

000 Ü'Ä0.008=Ü"Ã0.2Ü`=0.2

답 0.2

7

000 Þ"Ã(-3)Þ`=-3‌ ^답 -3

8

000 ß"Ã(-1)ß`=ß"Å1ß`=1‌ ^답 1

9

000 Ü®É-;2¥7;=Ü¾¨{-;3@;}3`=-;3@;

답 -;3@;

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(3)

30

00 (3^'4)^'25=3^'Ä100=3Ú`â`

답 3Ú`â`

31

00 8^{- '3}⁶_2^'3²‌‌=(2Ü`)^{- '3}⁶_2^'3²=2^{- '3}²_2^'3²=2^{- '3}^{2 +}^'3²=1 답 1

32

00 4^'2_4^'18Ö4^'8=4^'2+'18-'8=4^2'2=2^4'2

답 2^4'2

33

00 (4^'6¹_3^®;3@;)^'3=(2^'6²_3^'2^'3)^'3=2^'2_3^'2=6^'2

답 6^'2

34

00 (x^;2!;+y^;2!;)(x^;2!;-y^;2!;)=(x^;2!;)Û`-(y^;2!;)Û`=x-y

답 x-y

35

00 (x^;3!;+y^;3!;)(x^;3@;-x^;3!;y^;3!;+y^;3@;)=(x^;3!;)Ü`+(y^;3!;)Ü`=x+y 답 x+y

본문 8~12쪽

유형

^익^/^히^/^기

36

00 ㄱ.‌27의‌세제곱근‌중‌실수인‌것은‌Ü'¶27=Ü"Å3Ü`=3이다.

‌ (거짓)

ㄴ.‌‌‌'4=2의‌세제곱근‌중‌실수인‌것은‌Ü'2이다.‌(거짓) ㄷ.‌‌‌16의‌네제곱근‌중‌실수인‌것은‌ÑÝ'16=ÑÝ"Å2Ý`=Ñ2이다.‌

‌ (참)

ㄹ.‌‌‌'81=9의‌네제곱근을‌x라‌하면‌ ‌

xÝ`=9이므로‌xÝ`-9=0,‌(xÛ`+3)(xÛ`-3)=0‌ ‌ x=Ñ'3i‌또는‌x=Ñ'3의‌4개이다.‌(참)

따라서‌옳은‌것은‌ㄷ,‌ㄹ이다.

답 ㄷ, ㄹ

37

00 -64의‌세제곱근‌중‌실수인‌것은‌

Ü'Ä-64=Ü"Ã(-4)Ü`=-4의‌1개이므로‌a=1

5의‌네제곱근‌중‌실수인‌것은‌-Ý'5,‌Ý'5의‌2개이므로‌b=2

∴‌ab=2

답 2

38

00 ①‌25의‌제곱근을‌x라‌하면‌xÛ`=25에서‌

‌ xÛ`-25=0,‌(x+5)(x-5)=0‌ ‌ ∴‌x=Ñ5

②‌‌‌81의‌네제곱근을‌x라‌하면‌xÝ`=81에서‌

‌ xÝ`-81=0,‌(xÛ`+9)(xÛ`-9)=0

‌ (xÛ`+9)(x+3)(x-3)=0

‌ ∴‌x=Ñ3i‌또는‌x=Ñ3‌‌

‌ 이때‌81의‌네제곱근‌중‌실수인‌것은‌-3,‌3이다.

③‌제곱근‌9는‌'9=3이다.

④‌-1의‌제곱근을‌x라‌하면‌xÛ`=-1에서‌x=Ñi

⑤‌‌‌-27의‌세제곱근을‌x라‌하면‌xÜ`=-27에서‌ ‌ xÜ`+27=0, (x+3)(xÛ`-3x+9)=0‌ ‌

∴‌x=-3‌또는‌x= 3Ñ3'3i2 ‌ ‌

이때‌-27의‌세제곱근‌중‌실수인‌것은‌-3이다.

따라서‌옳은‌것은‌⑤이다.

답 ⑤

39

00 ①‌a<0일‌때,‌(Ü'Ä-a)Ü`=-a이다.‌

②‌(-2)Û`=4의‌제곱근은‌Ñ2이다.‌

③‌‌‌'¶256="Å2¡`=2Ý`이므로‌'¶256의‌네제곱근을‌x라‌하면‌‌ ‌ xÝ`=2Ý`에서‌xÝ`-2Ý`=0,‌(xÛ`+2Û`)(xÛ`-2Û`)=0‌ ‌ (xÛ`+2Û`)(x+2)(x-2)=0‌ ‌

④‌‌‌n이‌짝수이고‌a>0일‌때,‌xÇ`=a를‌만족시키는‌실수‌x의‌값은‌‌

Ç‌'a,‌-Ç‌'a의‌2개이다.‌

⑤‌‌‌n이‌홀수일‌때,‌-3의‌n제곱근‌중‌실수인‌것은‌ ‌ Ç‌'¶-3=-Ç '3이다.‌

따라서‌옳은‌것은‌⑤이다.‌

답 ⑤

40

00 ①‌Ü'2_Ü'4=Ü'8=Ü"Å2Ü`=2

②‌Ü"Ã2_Ü'64=Ü"Ã2_Ü"Å2ß`=Ü"Ã2_2Û`=Ü"Å2Ü`=2

③‌Ü'Ä-27

Ü'8 = Ü"Ã(-3)Ü`

`Ü"Å2Ü` = -32 =-;2#;

④‌{Ü'5_ 1'5 }6`=(Ü'5‌)ß`_{ 1'5 }6`=5Û`_ 15Ü`=;5!;

⑤‌"Ã2_Ü'4ÖÜ"Ã4'2="ÃÜ "2Ü`_4ÖÜ "Ã"4Û`_2=ß"Å2Þ`Öß"Å2Þ`=1 따라서‌옳지‌않은‌것은‌④이다.‌

답 ④

41

00 ⑴‌a='32ÖÝ'4='32ÖÝ"Å2Û`='32Ö'2='16=4

‌ b=Ü"Ã'64=ß'64=ß"Å2ß`=2

‌ ∴‌;bA;=;2$;=2

⑵‌Ú`Û"Ã2aÞ`bÝ`_Ý"Ã2abÛ`Öß"Ã4aÜ`b‌‌= Ú`Û"Ã2aÞ`bÝ`_Ú`Û"Ã2Ü`aÜ`bß`

Ú`Û"Ã4Û`aß`bÛ` ‌ ‌

=Ú`Û¾¨ 16a¡`bÚ`â`16aß`bÛ`=Ú`Û"aÛ`b¡`=ß"abÝ`

답 ⑴ 2 ⑵ ß "abÝ`

42

00 Ý7 9"Å2Þ`Ü'3_ß7 9'3 Ç "Å2á` ‌‌=Ý""Å2Þ`

Ý"Ü'3_ ß"'3

ß"Ç "Å2á` = ¡"Å2Þ`Ú`Û'3_ Ú`Û'3`

ß` Ç "Å2á` ‌ ‌

= ¡"Å2Þ`

ß` Ç "Å2á`

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(4)

단계 채점요소 배점

 주어진‌네‌수를‌같은‌거듭제곱근으로‌나타내기 50‌%

‌ a,‌b의‌값‌구하기 30‌%

‌ aÚ`Û`+bÚ`Û`의‌값‌구하기 20‌%

47

00 A="Ã2_Ü'3=Á°Ü"Å2Ü`_Ü'3="ÃÜ'24=ß'24 B=Ü"Ã3'2=ÜÁ°"Å3Û`_'2=Ü"Ã'18=ß'18

C=Ü"Ã2'3=ÜÁ°"Å2Û`_'3=Ü"Ã'12=ß'12 따라서‌ß'12<ß'18<ß'24이므로 C<B<A

답 ⑤

48

00 _{:ª5¦:}^;2!;_[{;1ª2¦5;}^-;3!;]^;2#;‌‌={ 275 }

;2!;_{;1ª2¦5;}^-;2!;‌ ‌

={ 275 }

;2!;_{:Á2ª7°:}^;2!;‌ ‌

={ 275 _:Á2ª7°:}

;2!;‌ ‌

=25¹²=5

답 5

49

00 aÑ¡`_(aÑÜ`)ÑÛ`ÖaÑÞ`‌‌=aÑ¡`_aß`ÖaÑÞ`‌ ‌

=a^-8+6-(-5)=aÜ`

∴‌k=3

답 3

50

00 ⑴‌27â`+{;3!;}- 3`=1+3Ü`=1+27=28

⑵‌3ÑÚ`â`+3Ú`Û`

3Ú`â`+3ÑÚ`Û`‌‌= 1 3Ú`â`+3Ú`Û`

3Ú`â`+ 13Ú`Û`

= 3Û`Û`+1

3Ú`â`

3Û`Û`+1 3Ú`Û`

= 3Ú`Û`(3Û`Û`+1) 3Ú`â`(3Û`Û`+1)‌= 3Ú`Û`

3Ú`â`‌‌‌

=3Û`=9

답 ⑴ 28 ⑵ 9

51

00 ㄱ.‌2^;3!;_2^;6!;=2^;3!;±^;6!;=2^;2!;='2‌(참) ㄴ.‌(9ÑÛ`)^;4!;=(3ÑÝ`)^;4!;=3ÑÚ`=;3!;‌(참) ㄷ.‌{(-3)Û`}^;2#;=(3Û`)^;2#;=3Ü`=27‌(거짓) ㄹ.‌('2)^2'2={('2)Û`}^'2=2^'2‌(거짓) 따라서‌옳은‌것은‌ㄱ,‌ㄴ이다.

답 ②

52

00 (주어진‌식)‌‌=a¡`Öa^2'3Ö(a^3-'3)Û`=a^8-2'3Öa^6-2'3‌‌

=a8-2'3-6+2'3=aÛ`=aû`

∴‌k=2

답 2

¡"Å2Þ`

ß` Ç "Å2á` =¡'4=¡"Å2Û`‌에서‌ß`Ç‌"Å2á`= ¡"Å2Þ`"Å2Û` ¡ =¡"Å2Ü`=Û`Ý"Å2á`

따라서‌6n=24이므로‌n=4

답 4

43

00 Ü¾¨ Ý'aÞ'aÖÝ¾¨ Ü'aÞ'a_Þ¾¨ Ü'aÝ'a‌‌= Ü"Ý'a Ü"ÃÞ'a_ Ý"Þ'a

Ý"ÃÜ'a_ Þ"Ü'a Þ"ÃÝ'a‌ ‌

= Ú`Û'a Ú`Þ'a_ Û`â'a

Ú`Û'a_ Ú`Þ'a Û`â'a=1

답 1

44

00 A="'5=Ý'5,‌B=Ü'3,‌C="ÃÜ'10=ß'10에서 4,‌3,‌6의‌최소공배수가‌12이므로

A=Ý'5=Ú`Û"Å5Ü`=Ú`Û'¶125 B=Ü'3=Ú`Û"Å3Ý`=Ú`Û'81 C=ß'10=Ú`Û"10Û`=Ú`Û'¶100

따라서‌Ú`Û'81<Ú`Û'¶100<Ú`Û'¶125이므로‌

B<C<A

답 ④

45

00 A=Ü®;4!;,‌B=Ý®;6!;,‌C=Ü¾¨®É;1Á7;=ß®É;1Á7;‌에서 3,‌4,‌6의‌최소공배수가‌12이므로

A=Ü®;4!;=Ú`Û¾¨{;4!;}4`=Ú`Û®É;25!6;

B=Ý®;6!;=Ú`Û¾¨{;6!;}3`=Ú`Û®É;21!6;

C=ß®É;1Á7;=Ú`Û¾¨{;1Á7;}2`=Ú`Û®É;28!9;

따라서‌Ú`Û®É;28!9;<Ú`Û®É;25!6;<Ú`Û®É;21!6;이므로 C<A<B

답 ⑤

46

00 '2,‌Ü'3,‌Ý'5,‌Ü"'7=ß'7에서‌2,‌3,‌4,‌6의‌최소공배수가‌

12이므로 '2=Ú`Û"Å2ß`=Ú`Û'64 Ü'3=Ú`Û"Å3Ý`=Ú`Û'81 Ý'5=Ú`Û"Å5Ü`=Ú`Û'¶125 ß'7=Ú`Û"Å7Û`=Ú`Û'49



∴‌Ú`Û'49<Ú`Û'64<Ú`Û'81<Ú`Û'¶125

따라서‌가장‌큰‌수는‌Ú`Û'¶125,‌가장‌작은‌수는‌Ú`Û'49이므로‌

a=Ú`Û'Ä125,‌b=Ú`Û'49



∴‌aÚ`Û`+bÚ`Û`=125+49=174

 답 174

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(5)

58

00 ⑴‌ÚÞaÛ`_Á°a_Ü"ÅaÝ`‌‌="ÅaÛ`_"'a_ÚÞÁ°Ü"ÅaÝ`‌‌

="ÅaÛ`_Ý'a_Ú`Û"ÅaÝ`‌ ‌

=a_a¹⁴_a^;3!;‌ ^‌

=a^{1+ 14+;3!;}=a^;1!2(;

‌ Ü7 9Ý"aÇ `

"aÞ`= ÜÁ°Ý"aÇ ` ÜÁ°"aÞ`= Ú`Û"aÇ `

ß"aÞ`‌= a^;1÷2;

a^;6%;=a^{12 -;6%;}ⁿ =a^n-10¹²

‌ 따라서‌a^;1!2(;=a^n-10¹² 이므로‌

‌ 19=n-10‌ ‌ ∴‌n=29

⑵‌Ç‌Á°27_Ü"Ã9_Ý'3‌‌=Ç‌'27_Ç‌"Ü'9_Ç‌Á°Ü"Ý'3‌ ‌

=Ç‌"3Ü`_³ⁿ"3Û`_¹²ⁿ'3‌‌

=3³ⁿ_3³ⁿ²_3^12n‌¹ ‌ ‌

=3³^{n +}^{3n +}² ^12n‌¹ ‌ ‌

=3^12n‌⁴⁵=3⁴ⁿ¹⁵

‌ Ý"Å3Ü`=3^;4#;이므로‌15

4n =;4#;‌ ‌ ∴‌n=5

답 ⑴ 29 ⑵ 5

59

00 A‌‌=Ü®É4'4_ 4

Ý'4 ={4'4_ 4 Ý'4 }

;3!;‌ ‌

=(4^1+;2!;_4^1-;4!;)¹³=(4^;2#;+;4#;)^;3!;‌ ‌

=(4^;4(;)¹³=4^;4#;=(2Û`)^;4#;=2^;2#;

따라서‌AÇÇ`,‌즉‌(2³²)Ç`‌이‌정수가‌되도록‌하는‌자연수‌n의‌최솟값

은‌2이다.‌ 답 2

60

00 5¡`=a,‌8ß`=b에서‌5=a^;8!;,‌8=b^;6!;이므로 200Ú`â`=(5Û`_8)Ú`â`={(a^;8!;)Û`_b^;6!;}Ú`â`=a^;2%;b^;3%;

답 ④

61

00 a=Ü'2,‌b=Ý'3에서‌aÜ`=2,‌bÝ`=3이므로‌

Ú`Û"Å6à`=Ú`Û"Ã(2_3)à`=Ú`Û"Ã(aÜ`bÝ`)à`=(aÜ`bÝ`)^;1¦2;=a^;4&;b^;3&;

답 ①

62

00 a=25Û`=5Ý`에서‌a^;4!;=5이므로‌

125Ü`=(5Ü`)Ü`=5á`=(a^;4!;)á`=a^;4(;=aû`

∴‌k=;4(;‌ ^답;4(;

63

00 aÝ`=2,‌bÚ`â`=8에서‌a=2^;4!;,‌b=8^;1Á0;이므로 (ß"ÃaÛ`bÞ`‌)û`‌‌=(aÛ`bÞ`)^;6K;={(2^;4!;)Û`_(8^;1Á0;)Þ`}^k⁶‌ ‌

=(2^;2!;_8^;2!;)^k⁶‌ ‌

=(2^;2!;_2^;2#;)^k⁶‌ ‌

=(2Û`)^k⁶=2^;3K;

53

00 (a^-;3!;b^;2!;)^;2!;_(a^;3$;b^-;4#;)^-1

=a^-;6!;_b^;4!;_a^-;3$;_b^;4#;

=a^-;6!;-;3$;_b^;4!;+;4#;=a^-;2#;b

= ba'a= b'a aÛ`

답 ②

54

00 18³²_24^;3@;Ö9^-;4#;‌‌=(2_3Û`)^;2#;_(2Ü`_3)^;3@;Ö(3Û`)^-;4#;‌ ‌

=2³²_3Ü`_2Û`_3^;3@;_3^;2#;‌ ‌

=2³^{2 +2}_33+;3@;+;2#;‌ ‌

=2⁷²_3^:£6Á:

∴‌x=;2&;,‌y=:£6Á:‌ ‌

∴‌x+y=:ª3¤:

답:ª3¤:

55

00 _{_{`2Ú`Û` }}¹ ^;n!;=(2ÑÚ`Û`)^;n!;=2^{- 12}ⁿ

 2Ñ¹²ⁿ이‌정수이려면‌- 12n 가‌음이‌아닌‌정수이어야‌하므로

 정수‌n은‌-1,‌-2,‌-3,‌-4,‌-6,‌-12의‌6개이다.

 답 6

 { 12Ú`Û`‌}

;n!;

을‌지수법칙을‌이용하여‌간단히‌하기 30‌%

‌ { 12Ú`Û`‌}

;n!;이‌정수가‌되도록‌하는‌n의‌조건‌구하기 40‌%

‌ 정수‌n의‌개수‌구하기 30‌%

56

00 "Ý'a_Á°a"Ãa'a‌‌=¡'a_'a_"'a_Á°"'a‌ ‌

=¡'a_'a_Ý'a_¡'a‌ ‌

=a¹⁸_a^;2!;_a^;4!;_a^;8!;‌ ‌

=a¹8 +;2!;+;4!;+;8!;=aÚ`=a

답 ④

57

00 P='a‌Ü'a‌Ý'a=a^;2!;_a^;3!;_a^;4!;=a;2!;+;3!;+;4!;=a^;1!2#;

Q=aÁ°a"Åaû`‌‌=a_'a_Á°"Åaû`=aÚ`_a^;2!;_a^;4K;=a1+;2!;+;4K;=a^6+k⁴ P='§Q에서

a^;1!2#;=ÚÞa^6+k⁴ ,‌a^;1!2#;=a^6+k⁸

;1!2#;= 6+k8 ‌ ‌ ∴‌k=;3*; 답 ;3*;

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(6)

25Å`+10+25Ú`ÑÅ`=64

∴‌25Å`+25Ú`ÑÅ`=54‌ 답 54

70

00 _{{'x+ 1}_{'x }}Û`=x+;[!;+2에서‌9=x+;[!;+2

∴‌x+;[!;=7



또‌{x+;[!;}Û`=xÛ`+ 1xÛ`+2에서‌49=xÛ`+ 1xÛ`+2

∴‌xÛ`+ 1xÛ`=47



∴‌xÛ`+xÑÛ`+7

x+xÑÚ`+2=47+7 7+2 =6

 답 6

 x+;[!;의‌값‌구하기 40‌%

‌ xÛ`+ 1

xÛ`‌의‌값‌구하기 40‌%

‌ 주어진‌식의‌값‌구하기 20‌%

71

00 ⑴‌5^;2A;+5^-;2A;='10의‌양변을‌제곱하면

‌ 5`+2+5Ñ`=10‌ ‌ ∴‌5`+5Ñ`=8

‌ ∴‌5^3a-5^2a+5`

5^2a ‌‌=5`-1+ 15^a=5`+5Ñ`-1‌‌

=8-1=7

⑵‌x=3^;3!;-3^-;3!;의‌양변을‌세제곱하면‌

‌ xÜ`=3-;3!;-3(3^;3!;-3^-;3!;)

‌ xÜ`=;3*;-3x,‌3xÜ`=8-9x

‌ ∴‌3xÜ`+9x-8=0

‌ ∴‌3xÝ`+3xÜ`+9xÛ`+x‌‌=x(3xÜ`+9x-8)+3xÜ`+9x‌ ‌

=8

답 ⑴ 7 ⑵ 8

72

00 aÛ`Å`=10이므로 aÅ`-aÑÅ`

aÅ`+aÑÅ`의‌분모,‌분자에‌각각‌aÅ`‌을‌곱하면 aÅ`-aÑÅ`

aÅ`+aÑÅ`=aÅ`(aÅ`-aÑÅ`)

aÅ`(aÅ`+aÑÅ`)=aÛ`Å`-1

aÛ`Å`+1= 10-110+1 =;1»1;

답;1»1;

73

00 ^aÅ`+aÑÅ``_aÅ`-aÑÅ`= aÅ`(aÅ`+aÑÅ`)

aÅ`(aÅ`-aÑÅ`)= aÛ`Å`+1 aÛ`Å`-1=3 aÛ`Å`+1=3(aÛ`Å`-1)

2aÛ`Å`=4‌ ‌ ∴‌aÛ`Å`=2 답 2

따라서‌(ß"ÃaÛ`bÞ`‌)û`,‌즉‌2^;3K;이‌자연수가‌되도록‌하는‌자연수‌k는‌3의‌

배수이므로‌k의‌최솟값은‌3이다.

답 3

64

00 (a^;2!;-b^;2!;)(a^;2!;+b^;2!;)(a+b)

={(a^;2!;)Û`-(b^;2!;)Û`}(a+b)

=(a-b)(a+b)=aÛ`-bÛ`

답 aÛ`-bÛ`

65

00 {2^'2+('2‌)^'2}{2^'2-('2‌)^'2}

=(2^'2)Û`-{('2‌)^'2}Û`=2^2'2-{('2‌)Û`}^'2

=2^'2+'2-2^'2=2^'2_2^'2-2^'2

=2^'2(2^'2-1)

답 ①

66

00 (x^;3!;+x^-;3@;)Ü`+(x^;3!;-x^-;3@;)Ü`

={(x^;3!;)Ü`+3(x^;3!;)Û`x^-;3@;+3x^;3!;(x^-;3@;)Û`+(x^-;3@;)Ü`}

+{(x^;3!;)Ü`-3(x^;3!;)Û`x^-;3@;+3x^;3!;(x^-;3@;)Û`-(x^-;3@;)Ü`}

=2(x+3xÑÚ`)

위의‌식에‌x=2를‌대입하면‌

2{2+;2#;}=4+3=7

답 7

67

00 ㄱ.‌(a^;4!;+b^;4!;)(a^;4!;-b^;4!;) =(a^;4!;)Û`-(b¹⁴)Û`‌ ‌

=a^;2!;-b¹^{2 ‌} ^‌

='a-'b‌‌(참)

ㄴ.‌(a^;2!;+a^{- 12}+1)(a^;2!;+a^-;2!;-1)‌‌=(a^;2!;+a^-;2!;)Û`-1Û`‌ ‌

=a+2+a^-1-1‌ ‌

=a+ 1a +1‌(참) ㄷ.‌(Ü'2+1)(Ü'4-Ü'2+1)=(Ü'2‌)Ü`+1Ü`=3‌(거짓) 따라서‌옳은‌것은‌ㄱ,‌ㄴ이다.

답 ㄱ, ㄴ

68

00 a^;3!;+a^-;3!;='5‌의‌양변을‌세제곱하면

a+aÑÚ`+3(a^;3!;+a^-;3!;)=('5 )Ü`

∴‌a+aÑÚ`‌‌=('5 )Ü`-3'5=2'5

답 ⑤

69

00 5Å`+5^1-x=8의‌양변을‌제곱하면 5Û`Å`+2´5Å`´5^1-x+5^2(1-x)=64

(5Û`)Å`+2´5^x+(1-x)+(5Û`)^1-x=64

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‌ 즉‌(k^;[!;)`´k^;]!;=(k^;z!;)Û`에서

‌ (2Ü`)`´3Û`=(2Û`´3)Û`,‌2Ü``´3Û`=2Ý`´3Û`

‌ 따라서‌3a=4이므로‌a=;3$;

⑵‌4Å`‌=5´`=10½`=k`(k>0)라‌하면‌xyz+0에서‌k+1

‌ 4Å`‌=k에서‌2^2x=k‌ ‌ ∴‌2=k^2x¹‌ yy‌㉠

‌ 5´`=k에서‌5=k^;]!;‌ yy‌㉡

‌ 10½`=k에서‌10=k^;z!;‌ yy‌㉢

‌ ㉠_㉡Ö㉢을‌하면

‌ 2_5Ö10=k2x +;]!;-;z!;¹ ‌ ‌ ∴‌k2x +;]!;-;z!;¹ =1

‌ 그런데‌k+1이므로‌ 12x +;]!;-;z!;=0

답 ⑴‌;3$;‌ ⑵‌0

79

00 xy+0이므로‌x+y-2xy=0의‌양변을‌xy로‌나누면

;]!;+;[!;-2=0‌ ‌ ∴‌;[!;+;]!;=2‌ yy‌㉠

2^2x=4Å`=k에서‌4=k^;[!;‌ yy‌㉡

3^2y=9´`=k에서‌9=k^;]!;‌ yy‌㉢‌

㉡_㉢을‌하면‌36=k^;[!;+;]!;

㉠에서‌;[!;+;]!;=2이므로‌

kÛ`=36‌ ‌ ∴‌k=6‌(∵‌k>0)

답 6

80

00 1회‌확대‌복사할‌때마다‌글자‌크기가‌r배‌커진다고‌하면‌

5회째의‌복사본의‌글자‌크기는‌처음‌원본의‌글자‌크기의‌2배이 므로‌

rÞ`=2‌ ‌ ∴‌r=2^;5!;

8회째의‌복사본의‌글자‌크기는‌4회째의‌복사본의‌글자‌크기의‌

rÝ` 배이므로 rÝ`=(2^;5!;)Ý`=2^;5$;

따라서‌m=5,‌n=4이므로 m+n=9

답 9

81

00 도시의‌인구가‌매년‌일정한‌비율로‌증가하므로‌1년마다‌

인구‌수가‌r배가‌된다고‌하면‌

1995년‌말부터‌2015년‌말까지‌20년‌동안‌인구는 r‌²⁰=6760000Ö40000=169(배)

1995년‌말부터‌2005년‌말까지‌10년‌동안‌인구는 r‌¹⁰=(r‌²⁰)^;2!;=169^;2!;=13(배)

따라서‌2005년‌말의‌인구는

40000_13=520000(명)=52(만‌명)

답 52만명

74

00

3Å`-3ÑÅ`

3Å`+3ÑÅ`= 3Å`(3Å`-3ÑÅ`)3Å`(3Å`+3ÑÅ`)= 3Û`Å`-13Û`Å`+1= 9Å`-19Å`+1 =;3!;

3´9Å`-3=9Å`+1 2´9Å`=4‌ ‌ ∴‌9Å`=2

∴‌9Å`-9ÑÅ`=9Å`-(9Å`)ÑÚ`=2-2ÑÚ`=2-;2!;=;2#;

답 ④

75

00 2ß`Å`-2Ñß`Å`

2Û`Å`+2ÑÛ`Å`‌‌= 2Û`Å`(2ß`Å`-2Ñß`Å`)2Û`Å`(2Û`Å`+2ÑÛ`Å`)‌ ‌

= 2¡`Å`-2ÑÝ`Å`2Ý`Å`+1 =(2Ý`Å`)Û`-(2Ý`Å`)ÑÚ`

2Ý`Å`+1 ‌ ‌

=9-;3!;

3+1 =:Á6£:

답:Á6£:

본문 13쪽

유형

76

00 aÅ`=16=2Ý`에서‌a=2^;[$;

b´`=16=2Ý`에서‌b=2^;]$;

∴‌ab=2^;[$;´2^;]$;=2^;[$;+;]$;=2Ý`‌{;[!;+;]!;}=2Ü`

따라서‌4{;[!;+;]!;}=3이므로‌

;[!;+;]!;=;4#;

답 ;4#;

77

00 2Å`=a에서‌2=a^;[!;‌ yy‌㉠

3´`=a에서‌3=a^;]!;‌ yy‌㉡

5½`=a에서‌5=a^;z!;‌ yy‌㉢

㉠_㉡_㉢을‌하면 30=a;[!;+;]!;+;z!;

그런데‌;[!;+;]!;+;z!;=2이므로‌

aÛ`=30‌ ‌ ∴‌a='30 (∵‌a>0)

답'30

78

00 ⑴‌8Å`=9´`=12½`=k`(k>0)라‌하면‌xyz+0에서‌k+1

‌ 8^x=2Ü`Å`=k에서‌2Ü`=k^;[!;

‌ 9‌^y=3Û`´`=k에서‌3Û`=k^;]!;

‌ 12^z=(2Û`´3)½`=k에서‌2Û`´3=k^;z!;

‌ 이때‌;[A;+;]!;=;z@;이므로‌k^;[A;´k^;]!;=k^;z@;

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87

00 ß"Ã8aÜ`bÜ`_Ú`ß"Ã256aß`bÝ`Ö'¶4ab

=ß"Ã2ÜàÜ`bÜ`_Ú`ß"Ã2¡àß`bÝ`Ö"Ã2Ûàb

="Ã2ab_¡"Ã2Ý`aÜ`bÛ`Ö"Ã2Û`ab

= ¡"Ã2Ý`aÝ`bÝ`_¡"Ã2Ý`aÜ`bÛ`

¡"Ã2¡`aÝ`bÝ`

=¡¾¨ 2¡`aà`bß`2¡`aÝ`bÝ`

=¡"aÜ`bÛ`

답 ③

다른풀이‌ 거듭제곱근을‌유리수인‌지수로‌바꾸어‌계산할‌수도‌있다.

ß"Ã8aÜ`bÜ`_Ú`ß"Ã256aß`bÝ`Ö'¶4ab

=8^;6!;a^;2!;b^;2!;_256^;1Á6;a^;8#;b^;4!;Ö4^;2!;a^;2!;b^;2!;

=2^;2!;a^;2!;b^;2!;_2^;2!;a^;8#;b^;4!;Ö2a^;2!;b^;2!;

=2;2!;+;2!;-1_a;2!;+;8#;-;2!;_b;2!;+;4!;-;2!;

=2â`_a^;8#;_b^;4!;=a^;8#;b^;4!;

=a^;8#;b^;8@;=¡"ÃaÜ`bÛ`

88

00 Å '2_´'4=^x´"Å2´`_^x´"Å4Å`=^x´"2´`´4Å`=Ú`¡"Ã2^2x+y‌(∵‌xy=18) x>0,‌y>0이므로‌산술평균과‌기하평균의‌관계에‌의하여 2x+y‌‌¾2'Ä2xy‌ ‌

=2'36=12‌(단,‌등호는‌2x=y일‌때‌성립)

∴‌Å '2_´'4‌‌=Ú`¡"Ã2^2x+y‌ ^‌

¾Ú`¡"2Ú`Û`=Ü"Å2Û`=Ü'4‌

답 ③

89

00 A=Ü"Ã2'4=Ü"Ã'4_'4=Ü"Ã'16=ß'§16 B="Ã2_Ü'4="ÃÜ'8_Ü'4="ÃÜ'32=ß'§32 C=Ü"Ã3'3=Ü"Ã'9_'3=Ü"Ã'27=ß'§27 따라서‌ß'16<ß'27<ß'32이므로 A<C<B

답 ②

90

00 ①‌"ÃÜ'¶5_6=ß'30

②‌"Ã6_Ü'5=Á°Ü"6Ü`_Ü'5="Ü'Ä1080=ß'Ä1080

③‌"Ã5_Ü'6=Á°Ü"5Ü`_Ü'6="Ü'Ä750=ß'Ä750

④‌Ü"Ã5'6=ÜÁ°"5Û`_'6=Ü"'Ä150=ß'Ä150

⑤‌Ü"Ã6'5=ÜÁ°"6Û`_'5=Ü"'Ä180=ß'Ä180 따라서‌가장‌큰‌수는‌②이다.

답 ②

91

00 {;2Á7;}^;n$;=(3ÑÜ`)^;n$;=3Ñ¹²^n‌

16^-;n!;=(2Ý`)^-;n!;=2^-;n$;

82

00 t년‌후에‌반감기가‌300년인‌방사능‌물질의‌양‌m이‌ m16 이‌된다고‌하면‌

16 =m´{;2!;}m

;30T0;

,‌;1Á6;={;2!;}^;30T0;,‌{;2!;}Ý`={;2!;}^;30T0;

4=;30T0;‌ ‌ ∴‌t=1200 따라서‌1200년‌후이다.

답 1200년 후

본문 14~17쪽

꼭

^{나오는 문제}

시험에

83

00 -27의‌세제곱근‌중‌실수인‌것은‌Ü'Ä-27=-3의‌1개이 므로‌a=1

10의‌네제곱근‌중‌실수인‌것은‌-Ý'10,‌Ý'10의‌2개이므로‌b=2

∴‌a+b=3

답 ②

84

00 ㄱ.‌n이‌홀수이면‌xÇ`=a`(a<0)를‌만족시키는‌실수‌x 는‌Ç‌'a의‌1개이다.‌(참)

ㄴ.‌‌‌n이‌짝수이면‌3의‌n제곱근‌중‌실수인‌것은‌Ç‌'3,‌-Ç‌'3의‌2개 이다.‌(참)

ㄷ.‌‌‌(반례)‌n=2,‌a=2일‌때,‌'¶-2='2i이므로‌‌‌

'¶-2+-'2이다.‌(거짓)

ㄹ.‌‌‌81의‌네제곱근을‌x라‌하면‌xÝ`=81이므로‌‌ ‌ xÝ`-81=0,‌(xÛ`+9)(xÛ`-9)=0‌ ‌

∴‌x=Ñ3i‌또는‌x=Ñ3‌(참) 따라서‌옳은‌것은‌ㄱ,‌ㄴ,‌ㄹ이다.

답 ④

85

00 Ü'Ä-27+Ý'48

Ý'3 +Ü"Ã'64‌‌=Ü"Ã(-3)Ü`+Ý®É 483 +ß'64‌ ‌

=Ü"Ã(-3)Ü`+Ý"2Ý`+ßß"2ß`‌ ‌

=-3+2+2=1

답 ③

86

00 (ß'9-Ü'24-2_á'Ä-27‌)ß`

=(ß"3Û`-Ü"Ã2Ü`_3-2_á"Ã(-3)Ü`‌)ß`

=(Ü'3-2 Ü'3+2 Ü'3‌)ß`=(Ü'3‌)ß`

=3Û`=9

답 ②

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97

00 Ü'§x+ 1Ü'§x=4에서‌x^;3!;+xÑ^;3!;=4 위의‌식의‌양변을‌제곱하면‌

x^;3@;+2+xÑ^;3@;=16

∴‌x^;3@;+xÑ^;3@;=14

위의‌식의‌양변을‌제곱하면‌

x^;3$;+2+xÑ^;3$;=196

∴‌x^;3$;+xÑ^;3$;=194

∴‌Ü"ÅxÝ`+ 1Ü"xÝ`=x^;3$;+xÑ^;3$;=194

답 194

98

00 x=3^;3!;+3^-;3!;의‌양변을‌세제곱하면

xÜ`=3+3ÑÚ`+3(3^;3!;+3^-;3!;) xÜ`=3+;3!;+3x,‌3xÜ`=9+1+9x

∴‌3xÜ`-9x-10=0

∴‌3xÜ`-9x-8=(3xÜ`-9x-10)+2=0+2=2

답 2

99

00 2Å`+2ÑÅ`=4의‌양변을‌제곱하면 4Å`+2+4ÑÅ`=16

∴‌4Å`+4ÑÅ`=14

2Å`+2ÑÅ`=4의‌양변을‌세제곱하면 8Å`+8ÑÅ`+3(2Å`+2ÑÅ`‌)=64 8Å`+8ÑÅ`+3´4=64

∴‌8Å`+8ÑÅ`=52

∴‌8Å`+8ÑÅ`

4Å`+4ÑÅ`=;1%4@;=:ª7¤:

따라서‌m=7,‌n=26이므로‌

m+n=33

답 33

100

0 주어진‌식의‌분모,‌분자에‌각각‌aÚ`â`을‌곱하면‌

(주어진‌식)‌‌= aÚ`â`(aÞ`+aÝ`+aÜ`+aÛ`+a) aÚ`â`(aÑá`+aÑ¡`+aÑà`+aÑß`+aÑÞ`)‌ ‌

=aÚ`â`(aÞ`+aÝ`+aÜ`+aÛ`+a)

a+aÛ`+aÜ`+aÝ`+aÞ` ‌ ‌

=aÚ`â`

이때‌aÞ`=7이므로‌

aÚ`â`=(aÞ`)Û`=7Û`=49

답 49

101

0 ^3Å`-3ÑÅ`_3Å`+3ÑÅ`의‌분모,‌분자에‌각각‌3Å`‌을‌곱하면‌

3Å`-3ÑÅ`

3Å`+3ÑÅ`=3Å`(3Å`-3ÑÅ` )

3Å`(3Å`+3ÑÅ` )= 3Û`Å`-13Û`Å`+1= 9Å`-19Å`+1=k이므로 9Å`-1=k(9Å`+1),‌9Å`(1-k)=k+1

{ 127 }

;n$;과‌16^-;n!;이‌모두‌자연수가‌되려면‌n<0이고‌|n|이‌12와‌

4의‌공약수이어야‌하므로‌n의‌값은‌-1, -2, -4 따라서‌구하는‌합은‌-1-2-4=-7

답 -7

92

00 (Ü"Å5Þ`‌)^;4!;=(5^;3%;)^;4!;=5^;1°2;

따라서‌(Ü"Å5Þ`‌)^;4!;,‌즉‌5^;1°2;이‌어떤‌자연수‌x의‌n제곱근이면‌

x=(5^;1°2;)ⁿ=5^;1°2;n이므로‌5^;1°2;n은‌자연수이다.‌

즉‌자연수‌n`(n¾2)은‌`12의‌배수이므로‌두‌자리‌자연수‌n은‌

12,‌24,‌36,‌y,‌96의‌8개이다.

답 8

93

00 ⑴‌Ü"ÅaÞ`=ÝÁ°a_Ü"Åaû`‌에서‌a^;3%;=a^;4!;a^;1ð2;=a^;4!;+;1ð2;이므로 ;3%;=;4!;+;1ð2;,‌20=3+k

‌ ∴‌k=17

⑵‌Á°2_Ü"Ã2_Ý'2‌‌='2_"Ü'2_Á°Ü"Ý'2='2_ß'2_Û`Ý'2‌ ‌

=2;2!;+;6!;+ 124=2¹⁷²⁴

‌ ∴‌n=17‌

답 ⑴ 17 ⑵ 17

94

00 `a='2,‌b=Ü'3에서‌aÛ`=2,‌bÜ`=3이므로 Ú`Û'12‌=Ú`Û"Ã2Û`´3=Ú`Û"ÃaÝ`bÜ`‌ ‌

=(aÝ`bÜ`)^;1Á2;=a¹³b^;4!;‌

답 ①

95

00 2`=c에서‌(2`)ÑÚ`=cÑÚ`이므로‌{;2!;}`=;c!;

2º`=d에서‌(2ÑÚ`)Ñº`=d이므로‌{;2!;}^-b=d,‌{;2!;}^-2b=dÛ`

∴‌{;2!;}^a-2b‌‌={ 12 }^a_{;2!;}^-2b‌ ‌

=;c!;´dÛ`= dÛ`c

답 ②

96

00 (1+3Û`)(1+3)(1+3^;2!;)(1+3^;4!;)(1+3^;8!;)(1-3^;8!;)

=(1+3Û`)(1+3)(1+3^;2!;)(1+3^;4!;)(1-3^;4!;)

=(1+3Û`)(1+3)(1+3^;2!;)(1-3^;2!;)

=(1+3Û`)(1+3)(1-3)

=(1+3Û`)(1-3Û`)

=1-3Ý`=1-81=-80‌

답 -80

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106

0 바이러스의‌개체수가‌한‌시간‌후‌r배가‌된다고‌하면‌바이 러스‌한‌마리가‌8시간‌후에‌8마리로‌늘어나므로

r¡`=8‌ ‌

∴‌r‌¹⁶=(r‌⁸)Û`=8Û`=64

따라서‌바이러스‌한‌마리가‌16시간‌후에‌64마리로‌늘어난다.

답 64마리

107

0 _{¾¨'a_ a}_Ü_'a‌‌=(a^;2!;_a_a^-;3!;)^;2!;

=(a;2!;+1-;3!;)¹²‌ ‌

=(a^;6&;)¹²=a^;1¦2;



"Ã'a_Ü'a`

ÝÁ°Ü"ÅaÛ` =(a^;2!;_a^;3!;)^;2!;

(a^;3@;)^;4!; =a^;1°2;

a^;6!;=a^;4!;

 따라서‌

¾¨'a_ aÜ'aÖ "Ã'a_Ü'a`

ÝÁ°Ü"ÅaÛ` ‌‌=a^;1¦2;Öa^;4!;‌ ‌

=a^{;1¦2;- 14}

=a¹³ 이므로‌m=;3!;



답 ;3!;

 ¾¨'a_ aÜ'a‌‌간단히‌하기 40‌%

‌ "Ã'a_Ü'a

ÝÁ°Ü"ÅaÛ` ‌간단히‌하기 40‌%

‌ m의‌값‌구하기 20‌%

108

0 aÜ`=5,‌bÝ`=11,‌cß`=13에서‌

a=5^;3!;,‌b=11^;4!;,‌c=13^;6!;

 이므로‌(abc)Ç`=(5^;3!;_11^;4!;_13¹⁶)^n‌이‌자연수가‌되도록‌하는‌자 연수‌n은‌3,‌4,‌6의‌공배수,‌즉‌12의‌배수이다.‌

 따라서‌자연수‌n의‌최솟값은‌12이다.

 답 12

 a,‌b,‌c의‌값‌구하기 30‌%

‌ (abc)Ç`‌이‌자연수가‌되도록‌하는‌n의‌조건‌구하기 50‌%

‌ 자연수‌n의‌최솟값‌구하기 20‌%

∴‌9Å`= 1+k1-k

∴‌9Å`+9ÑÅ`‌‌= 1+k1-k +1-k

1+k =(1+k)Û`+(1-k)Û (1-k)(1+k) ‌ ‌

=2(1+kÛ`) 1-kÛ` ‌

답 ⑤

102

0 aÞ`Å`-aÑÞ`Å`

aÅ`-aÑÅ` 의‌분모,‌분자에‌각각‌aÅ`을‌곱하면 aÞ`Å`-aÑÞ`Å`

aÅ`-aÑÅ` ‌‌= aÅ` (a^5x-a^-5x)

a^x(a^x-a^-x) = aß`Å`-aÑÝ`Å`aÛ`Å`-1 = (aÛ`Å`)Ü`-(aÛ`Å`)ÑÛ`aÛ`Å`-1 ‌ ‌

=('2)Ü`-('2)ÑÛ`

'2-1 =2'2-;2!;

'2-1 = 7+3'22

답 7+3'2 2

103

0 5Å`=27에서‌5Å`=3Ü`이므로

5=3^;[#;‌ yy‌㉠

45´`=81에서‌45´`=3Ý`이므로

45=3^;]$;‌ yy‌㉡

㉠Ö㉡을‌하면

;4°5;= 3^;[#;

3^;]$;,‌;9!;=3^;[#;-;]$;

이때‌;9!;=3ÑÛ`‌이므로‌;[#;-;]$;=-2

답 -2

104

0 aÅ`=b´`=c½`=27에서‌

a=27‌^;[!;,‌b=27‌^;]!;,‌c=27‌^;z!;

∴‌abc=27‌;[!;+;]!;+;z!;=3Ü`‌{;[!;+;]!;+;z!;}=9 이때‌9=3Û`이므로‌

3{;[!;+;]!;+;z!;}=2‌ ‌ ∴‌;[!;+;]!;+;z!;=;3@;

답 ;3@;

105

0 8Å`=27´`=k`(k>0)라‌하면‌8=k^;[!;,‌27=k^;]!;

8=k^;[!;,‌27=k^;]!;‌을‌변끼리‌곱하면 8_27=k^;[!;_k^;]!;,‌216=k^;[!;+;]!;

6Ü`=kÜ`‌{∵‌;[!;+;]!;=3}‌ ‌ ∴‌k=6

8Å`=6이므로‌(2Ü`)Å`=6,‌(2Å`)Ü`=6‌ ‌ ∴‌2Å`=Ü'6 27´`=6이므로‌(3Ü`)´`=6,‌(3´`)Ü`=6‌ ‌ ∴‌3´`=Ü'6

∴‌(2Å`+3´`)Ü` =(Ü'6+Ü'6‌)Ü`=(2_Ü'6‌)Ü`‌ ‌

=2Ü`_6=48

답 48

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111

0 이차방정식‌xÛ`+2kx+6=0의‌두‌근이‌a,‌b이므로‌근 과‌계수의‌관계에‌의하여‌

a+b=-2k,‌ab=6

∴‌aÑÚ`-bÑÚ`

aÑÛ`-bÑÛ` ‌‌= aÑÚ`-bÑÚ`

(aÑÚ`+bÑÚ`)(aÑÚ`-bÑÚ`)‌ ‌

= 1

aÑÚ`+bÑÚ`= 1

;!;+;º!;‌ ‌

= aba+b =-;k#;

따라서‌-;k#;=;2¢5;이므로‌k=-:¦4°:

답 -:¦4°:

112

0 2`=x,‌2º`=y,‌2`=z라‌하면

xyz=2`‌2º`2`=2^a+b+c=2ÑÚ`=;2!;‌‌‌ yy‌㉠

x+y+z=2`+2º`+2`=:Á4£:

또한‌;[!;+;]!;+;z!;=2Ñ`+2Ñº`+2Ñ`=:Á2Á:이므로

;[!;+;]!;+;z!; = xy+yz+zxxyz ‌ ‌

=2(xy+yz+zx)‌(∵‌㉠)‌ ‌

= 112

∴‌xy+yz+zx=:Á4Á:

∴‌4`+4º`+4`‌=xÛ`+yÛ`+zÛ`‌ ‌

=(x+y+z)Û`-2(xy+yz+zx)‌‌

={ 134 }2`-2´11 4

= 8116

답;1*6!;

113

0 ㄱ.‌ f(10,‌2018)=2,‌ f(10,‌2017)=1,‌

f(-10,‌2017)=1이므로

f(10,‌2018)=f(10,‌2017)+f(-10,‌2017)‌(참)

ㄴ.‌‌‌(반례)‌a=0,‌n=2이면‌ f(0,‌5)=1,‌ f(0,‌4)=1이므로‌ ‌ f(0,‌5)+f(0,‌4)=2+3‌(거짓)

ㄷ.‌‌‌f('3,‌4)=2,‌ f(Ü'¶-6,‌7)=1,‌ f(-Ý'8,‌6)=0이므로‌ ‌ 4‌f('3,‌4)+3‌f(Ü'¶-6,‌7)+2‌f(-Ý'8,‌6)‌ ‌

=4´2+3´1+2´0=11‌(참) 따라서‌옳은‌것은‌ㄱ,‌ㄷ이다.

답 ③

109

0 x^;2!;+xÑ^;2!;=2'2의‌양변을‌제곱하면‌

x+2+xÑÚ`=8‌ ‌

∴‌x+xÑÚ`=6

 x^;2!;+xÑ^;2!;=2'2의‌양변을‌세제곱하면

x^;2#;+xÑ^;2#;+3(x^;2!;+xÑ^;2!;)=16'2 x^;2#;+xÑ^;2#;+6'2=16'2‌ ‌

∴‌x^;2#;+xÑ^;2#;=10'2



∴‌ x^;2#;+xÑ^;2#;

x+xÑÚ`+14= 10'26+14 ='2 2

 답 '2

2

 x+xÑÚ`의‌값‌구하기 40‌%

‌ x^;2#;+x^-;2#;의‌값‌구하기 40‌%

‌ x^;2#;+x^-;2#;

x+xÑÚ`+14‌의‌값‌구하기 20‌%

110

0 aÑÜ`Å`+aÜ`Å`

aÑÅ`+aÅ` =3의‌좌변의‌분모,‌분자에‌각각‌aÑÅ`‌을‌곱하면‌

aÑÝ`Å`+aÛ`Å`

aÑÛ`Å`+1 =3

 이때‌aÑÛ`Å`=t‌(t>0)라‌하면‌

tÛ`+;t!;

t+1 =3에서‌tÛ`+;t!;=3t+3 양변에‌t를‌곱하여‌정리하면‌

tÜ`-3tÛ`-3t+1=0,‌(t+1)(tÛ`-4t+1)=0

∴‌t=2Ñ'3 (∵‌t>0)



∴‌aÑÛ`Å`=2Ñ'3

 답 2Ñ'3

 등식의‌좌변의‌분모,‌분자에‌각각‌aÑÅ`‌을‌곱하여‌정리하기 30‌%

‌ aÑÛ`Å`=t‌(t>0)로‌놓고‌방정식을‌풀기 50‌%

‌ aÑÛ`Å`의‌값‌구하기 20‌%

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02 로그

^Ⅰ.^{지수함수와 로그함수}

본문 19쪽

교과서 문제

^정^/^복^/^하^/^기

114

0 답 4=log£`81

115

0 답 -3=log;3!;`27

116

0 logª`x=3에서‌x=2Ü`=8

답 8

117

0 log_;3!;`x=-2에서‌x={;3!;}ÑÛ`=9

답 9

118

0 log®`16=4에서‌xÝ`=16

∴‌x=16^;4!;=(2Ý`)^;4!;=2

답 2

119

0 log®`2=4에서‌xÝ`=2

∴‌x=2^;4!;=Ý'2

답 Ý'2

120

0 진수의‌조건에서‌x+1>0‌ ‌ ∴‌x>-1

답 x>-1

121

0 밑의‌조건에서‌x-5>0, x-5+1 즉 x>5, x+6

∴‌5<x<6‌또는‌x>6

답 5<x<6 또는 x>6

122

0 3`logª`4+2`logª`'2 =3`logª`2Û`+2`logª`2^;2!;

=6`logª`2+logª`2

=6+1

=7

답 7

123

0 log£`24+3`log£`3 2

=log£ (2Ü`´3)+3(log£`3-log£`2)

=log£`2Ü`+log£`3+3`log£`3-3`log£`2‌ ‌

=3`log£`2+1+3-3`log£`2‌ ‌

=4‌ 답 4

124

0 logª`18-2`logª`6

=logª (2´3Û`)-2`logª (2´3)

=logª`2+2`logª`3-2(logª`2+logª`3)

=1+2`logª`3-2-2`logª`3

=-1

답 -1

125

0 logÁ¼`12 =logÁ¼ (2Û`´3)

=2`logÁ¼`2+logÁ¼`3

=2a+b

답 2a+b

126

0 logÁ¼`;2¢7; =logÁ¼` 2Û`3Ü`

=2`logÁ¼`2-3`logÁ¼`3

=2a-3b

답 2a-3b

127

0 log£`16=logÁ¼`16

logÁ¼`3 =logÁ¼`2Ý`

logÁ¼`3=4`logÁ¼`2 logÁ¼`3 = 4ab

답 4a b

128

0 log¤`9 =logÁ¼`9

logÁ¼`6= logÁ¼`3Û`

logÁ¼ (2´3)

= 2`logÁ¼`3

logÁ¼`2+logÁ¼`3= 2ba+b

답 2b a+b

129

0 logª¦`81=log^3Ü``3Ý`=;3$;`log£`3=;3$;

답 ;3$;

130

0 log¢`;8!;=log2Û``2ÑÜ`=-;2#;`logª`2=-;2#;

답 -;2#;

131

0 3^log£`10=10^log£`3=10

답 10

132

0 log£`2´logª`9=logÁ¼`2

logÁ¼`3 ´logÁ¼`3Û`

logÁ¼`2=2`logÁ¼`3 logÁ¼`3 =2

답 2

133

0 log`1000=log`10Ü`=3

답 3

134

0 log`;10!0;=log`10ÑÛ`=-2

답 -2

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(13)

135

0 log`0.001=log`10ÑÜ`=-3

답 -3

136

0 log`Ü'Ä100=log`Ü"10Û`=log`10^;3@;=;3@;

답;3@;

137

0 답 0.7101

138

0 답 0.7007

139

0 log`534 =log (5.34_10Û`)

=log`5.34+log`10Û`

=0.7275+2

=2.7275

답 2.7275

140

0 log`0.0534 =log`(5.34_10ÑÛ`)

=log`5.34+log`10ÑÛ`

=0.7275-2

=-1.2725

답 -1.2725

141

0 log`48.2 =log`(4.82_10)

=log`4.82+log`10

=0.6830+1=1.6830

∴‌정수‌부분：1,‌소수‌부분：0.6830

답 정수부분：1, 소수부분：0.6830

142

0 log`4820 =log`(4.82_10Ü`)

=log 4.82+log`10Ü`

=0.6830+3=3.6830

∴‌정수‌부분：3,‌소수‌부분：0.6830

답 정수부분：3, 소수부분：0.6830

143

0 log`0.482 =log`(4.82_10ÑÚ`)

=log 4.82+log`10ÑÚ`

=-1+0.6830

∴‌정수‌부분：-1,‌소수‌부분：0.6830

답정수부분：-1, 소수부분：0.6830

144

0 log`0.0482 =log`(4.82_10ÑÛ`)

=log`4.82+log`10ÑÛ`

=-2+0.6830

∴‌정수‌부분：-2,‌소수‌부분：0.6830

답정수부분：-2, 소수부분：0.6830

145

0 답 3

146

0 답 -3

147

0 log`x=1.7348에서‌log`5.43과‌소수‌부분이‌같으므로‌

x는‌5.43과‌숫자의‌배열이‌같고,‌정수‌부분이‌1이므로‌정수‌부분 이‌두‌자리인‌수이다.

∴‌x=54.3

답 54.3

148

0 log`x=4.7348에서‌log`5.43과‌소수‌부분이‌같으므로‌

x는‌5.43과‌숫자의‌배열이‌같고,‌정수‌부분이‌4이므로‌정수‌부분 이‌다섯‌자리인‌수이다.

∴‌x=54300

답 54300

149

0 log`x=-0.2652=-1+0.7348에서‌log`5.43과‌소수‌

부분이‌같으므로‌x는‌5.43과‌숫자의‌배열이‌같고,‌정수‌부분이‌

-1이므로‌소수점‌아래‌첫째‌자리에서‌처음으로‌0이‌아닌‌숫자가‌

나타난다.

∴`x=0.543

답 0.543

150

0 log`x=-2.2652=-3+0.7348에서‌log`5.43과‌소수‌

부분이‌같으므로‌x는‌5.43과‌숫자의‌배열이‌같고,‌정수‌부분이‌

-3이므로‌소수점‌아래‌셋째‌자리에서‌처음으로‌0이‌아닌‌숫자가‌

나타난다.

∴‌x=0.00543

답 0.00543

본문 20~27쪽

유형

^익^/^히^/^기

151

0 log_'3`a=4에서 a=('3`)Ý`=(3^;2!;)Ý`=3Û`=9 log_;9!;`b=-;2!;에서 b={;9!;}^-;2!;=(3ÑÛ`)^-;2!;=3

∴‌ab=9´3=27

답 ②

152

0 ⑤‌7 ^;2!;='7 HjK log¦`'7=;2!;

답 ⑤