1. 次の対数の値を求めよ。
1 125
( ) log5 ( ) log102 1
10000 3
( ) log3 3 ( ) log104 1
5 9 ( ) log
3 ( ) log5 256
7 9
( ) log27 ( ) log18 8
2 9
( ) log16 1 2
2. 次の式を簡単にせよ。
1 8+ 64
( ) log3 log3 ( ) log5 log52 81- 243
3 45- 75
( ) log3 log3 ( )3 3log5180-2log51080
5 - ( ) log65
24 log615
2 ( ) log4 24 log4 186 -
3. 次の式を簡単にせよ。
1 25+ 5
( ) log3 log9 ( ) log4 log82 7+ 49
3 8- 4
( ) log9 log 27
4. 次の式を簡単にせよ。
1 4⋅ 5
( ) log3 log4 ( ) log5 log52 27÷ 243
3 ( 4+ 32 ( 5- 25
( ) log5 log25 ) log2 log 2 )
5. 次の式の値を求めよ。
1 6
( ) -log67 ( )2 32-log32
6. a =log52, b =log53 とするとき、次の式を a, b で表せ。
1 108
( ) log5 ( ) log52 30
3 18
( ) log25 ( ) log5 124
7. a =log212, b =log315 とするとき、次の式を a, b で表せ。
1 3+ 5
( ) log2 log3 ( ) log2 log32 3+ 5+3
©
ちゃんと理解したい⼈のための⾼校数学
@YouTube3
3
3
1. 次の対数の値を求めよ。
1 125 = 5 = 3
( ) log5 log5 3 ( ) log102 1 = 10 = -4
10000 log10 -4
3 = 3 =
( ) log3 3 log3 1 3 1
3
4 1 = 0 ( ) log10
5 9 = ( = 4
( ) log 3 log
3 3)4
6 = (5 = 5 =
( ) log5 25 log5 2) 1 3 log5
2 3 2
3
7 9 = = =
( ) log27 9 27 log3 log3
3 3 log3 2 log3 3
2
3 ( ) log18 8 = = = -3
2 log28 log21
2 2 2 log2 3 log2 -1 9
( ) log16 1 2
=log16 1 = 2
1 2 1
2log161 2
= ⋅1 = -
2 2
2 log2 -1
log2 4 1 8
2. 次の式を簡単にせよ。
1 8+ 64 ( ) log3 log3
=log3(2 ⋅2 =3 6) log32 = 99 2 log3
2 81- 243 ( ) log5 log5
=log581 = = 3 = - 3 243 log51
3 log5 -1 log5 3 45- 75
( ) log3 log3 =log345=
75 log33 =log3 log33- 5 5 = 1-log35
3 3 180-2 1080 ( ) log5 log5 =log5180 -3 log510802 = log5(2 ⋅3 ⋅5
(2 ⋅3 ⋅5 2 2 )3 3 3 )2 =log55 = 1 5 -
( ) log65 24 log615
2 =log6 5⋅
24 2 15
=log61 = 6 = -2 36 log6 -2
6 -
( ) log4 24 log4 18 =log4 24=
18 log4 4 3
=log44 - 3 = - 3 1
2 log4 1 2 1
2 1 2log4
3. 次の式を簡単にせよ。
1 25+ 5 ( ) log3 log9 =log35 +2 5
3 log3 log3 2
= 2log35+1 5 = 5 2log3 5
2log3
2 7+ 49 ( ) log4 log8 = 7+
2 log2 log2 2
7 2 log2 2 log2 3
=1 7+ 7 = 7
2log2 2 3log2 7
6log2
3 8- 4
( ) log9 log 27
= +
2 3 log3 3 log3 2
2
3 log3 2 log3
3 2
=3 2+ 2 = 2
2log3 4
3log3 17 6log3
©
ちゃんと理解したい⼈のための⾼校数学
@YouTube3
3
4. 次の式を簡単にせよ。
1 3⋅ 2 ( ) log2 log3
= log23⋅ = 2 = 1 2
3 log2 log2 log2
2 8÷ 16 ( ) log3 log3 =log32 ÷3 log324 = 3log32÷4log32 =3
4
3 ( 4+ 32 ( 5- 25
( ) log5 log25 ) log2 log 2 ) = log52 +2 2 5-
5 log5 5
log5 2 log2 5 2 log2 2 log2
1 2 = 2log52+5 2 5-4 5 2log5 (log2 log2 ) =9 2⋅ -3 5
2log5 ( log2 ) = -27 2⋅ = -
2log5 5 2 log5 log5
27 2
5. 次の式の値を求めよ。
1 6 = A ( ) -log67
log66-log67= A -log6 log67 = Alog6
log67-1 log= 6A A = 7-1 1=
7 ∴ 6-log67=1
7
2 3 = 3 ⋅3 = 9⋅3 ( ) 2+log32 2 log32 log32 3log32= A
log33log32= A log3 log3 log32 = A A = 2
∴ 32+log32= 9⋅2 = 18
6. a =log52, b =log53 とするとき、次の式を a, b で表せ。
1 108 ( ) log5 =log5(2 ⋅32 3) =log52 +2 log533 = 2log52+3log53 = 2a+3b
2 30 ( ) log5 =log5(2⋅3⋅5) =log5 log5 log52+ 3+ 5 = a+b+1
3 18 ( ) log25 = (2⋅3
5 log5 2)
log5 2 = (1 2+ 3
2log5 log5 2) = (1 2+2 3 = a+2b
2log5 log5 ) 1 2( )
4 ( ) log5 12 =log5( )12 1 3
=log5(2 ⋅32 ) 1 3 = (1 2 + 3
3log5 2 log5 ) = (21 2+ 3 = 2a+b
3 log5 log5 ) 1 3( )
©
ちゃんと理解したい⼈のための⾼校数学
@YouTube3
7. a =log212, b =log315 とするとき、次の式を a, b で表せ。
1 3- 5 ( ) log2 log3 =log212-
4 log315
=log2 log2 log3 log3 )12- 4-(3 15- 3 = a-b-1
2 3+ 5+3 ( ) log2 log3
=(log23+2 +) (log35+1) =(log2 log2 ) (log3 log3 )3+ 4 + 5+ 3 =log212+log315 = a+b
©
