Logarithmes décimaux

Exercices sur le logarithme
décimal
1. Soient a et b ∈ R∗+ .
Simplifier:
(a)
(b)
(c)
(d)
à r
!3
b2
a
log 0, 1 · a
a
b3
µ
¶3 µ −4 3 ¶−2
10a3 b−2
a b
√
√
log
4
2
3
a a b
100 b2 a
´3
³√
3
a4 b−2
0, 001
√ √
log
4
b3 a3
³ −3 4 √
´
3
10 a√ b
log 0,01a
2 a3 b2
2
2. Calculer:
(a) log 2 + log 5
(b) 2 log 5 + log 12 − log 3
3. Si log 2 = α, exprimer en fonction de α :
log 4; log 16; log 40; log 14 ; log 0, 2
4. Si log b = a avec b ∈ R∗+ , alors déterminer:
√
√
b
; log 1b ; log b; log b5 ; 2 log 3b + log 5 b − log 9
log 10b; log 100
5. Déterminer domf et simplifier f (x) si possible:
(a) f (x) = log(4 − 3x)
(b) f (x) = log(4 − x2 )
(2x − 3)3
2−x
4x − 1
(d) f (x) = log
x−3
(e) f (x) = log |5x − 1|
(c) f (x) = log
1
x3 − 6x2 + 11x − 6
x2 + 3x + 2
(1 + x2 )3
(g) f (x) = log p
√
x + 1 + x2
(f) f (x) = log
6. Résoudre dans R les équations suivantes:
(a) log x = 1
(b) log x = 3
(c) log x = −4
(d) log(x + 4) + log x = 0
(e) log(x + 3) + log(x + 5) = log 15
(f) log(x + 1) = 3 − log(1 − 2x)
(g) log(1 − x) − log(x + 1) = −2
(h) log(x + 1) + log(x − 1) = log 3 + 4 log 2
(i) log(x2 + 5x + 6) = log(x + 11)
(j) log(1 − 5x) − log(x + 1) = −1
7. Résoudre dans R les équations suivantes:
(a) (log x)2 − 3 log x − 4 = 0
(b) 2(log x)2 − log x + 1 = 0
(c) (log x)2 + log x − 12 = 0
8. Résoudre dans R les inéquations suivantes:
(a) log x >
1
2
(b) 2 log x 6 −3
(c) log |2x + 1| + log |x + 3| < 1
(d) log 24 + log(3 − x) < log(x + 1) + log(25x − 49)
(e) log(3x2 − x − 2) > log(6x + 4)
(f) log(x + 2) + log(x − 4) < 2 log(x − 1)
2
Corrigé
1.
(a) log10 0.1
à r
2
a
b2
a
!3
a
b3
à r
!3
2
b
a
= log10 0.1 + log10 a2
+ log10 3
a
b
r 3
b2
a
6
= −1 + log10 a + log10
+ log10 3
a
b
b2
3
= −1 + 6 log10 a + log10
+ log10 a − log10 b3
2
a
3
= −1 + 6 log10 a + 3 log10 b − log10 a + log10 a − 3 log10 b
2
11
= −1 +
log10 a
2
¶3 µ −4 3 ¶−2
10a3 b−2
a b
√
√
4
a a2 b3
100 b2 a
10a3 b−2
a−4 b3
√
= 3 log10 √
− 2 log10
4
a a2 b3
100 b2 a
(b) log10
µ
3
3
= 3 log10 10+3 log10 a3 +3 log10 b−2 −3 log10 a− log10 a2 − log10 b3 −
2
2
2
2
−4
3
2
2 log10 a − 2 log10 b + 2 log10 100 + log10 b + log10 a
4
4
9
= 3 + 9 log10 a − 6 log10 b − 3 log10 a − 3 log10 a − log10 b + 8 log10 a −
2
1
6 log10 b + 4 + log10 b + log10 a
2
23
31
=7+
log10 a −
log10 b
2
2
³√
´3
3
0.001
a4 b−2
√ √
(c) log10
4
b3 a3
√
√
√
3
4
= log10 0.001 + 3 log10 a4 b−2 − log10 b3 − log10 a3
1
1
= −3 + log10 a4 + 3 log10 b−2 − log10 b3 − log10 a3
2
4
3
= −3 + 4 log10 a − 6 log10 b −
= −3 +
(d) log10
Ã
13
4
log10 a −
15
log10 b
2
√ !
10−3 a4 3 b
√
0.01a2 a3 b2
3
3
log10 b − log10 a
2
4
√
√
= log10 10−3 + log10 a4 + log10 3 b − log10 0.01 − log10 a2 − log10 a3 b2
1
3
= −3 + 4 log10 a + log10 b + 2 − 2 log10 a − log10 a − log10 b
3
2
2
1
= −1 + log10 a − log10 b
2
3
2.
(a) log10 2 + log10 5
= log10 (2 · 5)
= log10 10
=1
(b) 2 log10 5 + log10 12 − log10 3
= log10 52 + log10 12 − log10 3
µ
¶
25 · 12
= log10
3
µ
¶
300
= log10
3
= log10 100
=2
3.
(a) log10 4
= log10 22
= 2 log10 2
= 2a
4
(b) log10 16
= log10 24
= 4 log10 2
= 4a
(c) log10 40
= log10 (4 · 10)
= log10 4 + log10 10
= 2a + 1
(d) log10
1
4
= log10 1 − log10 4
= 0 − 2a
= −2a
(e) log10 0.2
= log10 (0.1 · 2)
= log10 0.1 + log10 2
= −1 + a
4.
(a) log10 10b log10 10 + log10 b = 1 + a
b
100
= log10 b − log10 100
=a−2
(b) log10
(c) log10
1
b
= log10 1 − log10 b
=0−a
= −a
5
√
(d) log10 b
= 12 log10 b
= 12 a
(e) log10 b5
= 5 log10 b
= 5a
√
(f) 2 log10 3b + log10 5 b − log10 9
= 2 (log10 3 + log10 b) + 15 log10 b − log10 32
= 2 log10 3 + 2 log10 b + 15 log10 b − 2 log10 3
= 2a + 15 a
= 11
5
5.
(a) f (x) = log10 (4 − 3x)
x ∈ domf ⇔ 4 − 3x > 0 ⇔ x <
¤
£
domf = −∞; 43
4
3
¤
£
⇔ x ∈ −∞; 43
f (x) n’est pas simplifiable!
(b) f (x) = log10 (4 − x2 )
x ∈ domf ⇔ 4 − x2 > 0 ⇔ (2 − x)(2 + x) > 0 ⇔ x ∈ ]−2; 2[
domf = ]−2; 2[
f (x) = log10 [(2 − x)(2 + x)]
= log10 |2 − x| + log10 |2 + x|
6
(c) f (x) = log10
(2x − 3)2
2−x
(2x − 3)2
> 0 et 2 − x 6= 0 ⇔ x ∈ ]−∞, 2[ \
2−x
½ ¾
3
domf = ]−∞, 2[ \
2
x ∈ domf ⇔
½ ¾
3
2
(2x − 3)2
2−x
= log10 (2x − 3)2 − log10 |2 − x|
= 2 log10 |2x − 3| − log10 |2 − x|
f (x) = log10
(d) f (x) = log10
4x − 1
x−3
¸
·
4x − 1
1
x ∈ domf ⇔
> 0 et x − 3 6= 0 ⇔ x ∈ −∞;
∪ ]3; +∞[
x−3
4
·
¸
1
∪ ]3; +∞[
domf = −∞;
4
4x − 1
x−3
= log10 |4x − 1| − log10 |x − 3|
f (x) = log10
(e) f (x) = log10 |5x − 1|
x ∈ domf ⇔ |5x − 1| > 0 ⇔ x ∈ R\
½ ¾
1
domf = R\
5
f (x) n’est pas simplifiable!
7
½ ¾
1
5
(f) f (x) = log10
x3 − 6x2 + 11x − 6
x2 + 3x + 2
x3 − 6x2 + 11x − 6
> 0 et x2 + 3x + 2 6= 0
x2 + 3x + 2
(x − 1)(x − 2)(x − 3)
⇔
> 0 et (x + 2)(x + 1) 6= 0
(x + 1)(x + 2)
⇔ x ∈ ]−2; −1[ ∪ ]1; 2[ ∪ ]3; +∞[
domf = ]−2; −1[ ∪ ]1; 2[ ∪ ]3; +∞[
x ∈ domf ⇔
x3 − 6x2 + 11x − 6
x2 + 3x + 2
(x − 1)(x − 2)(x − 3)
= log10
(x + 2)(x + 1)
= log10 |(x − 1)(x − 2)(x − 3)| − log10 |(x + 2)(x + 1)|
= log10 |x − 1|+log10 |x − 2|+log10 |x − 3|−log10 |x + 2|−log10 |x + 1|
f (x) = log10
(g) f (x) = log10
¡
¢3
1 + x2
p
√
x + 1 + x2
¢3
¡
√
1 + x2
x ∈ domf ⇔ p
> 0 et x + 1 + x2 > 0 et 1 + x2 ≥ 0
√
2
x+ 1+x
√
2
⇔ x+ 1+x >0
√
⇔ 1 + x2 > −x
1er cas: x ∈ R+ :
√
1 + x2 > −x toujours vérifié ∀x ∈ R+
⇔ x ∈ R+
dom1 = R+
2e cas: x ∈ R∗− :
√
1½+ x2 > −x
1 + x2 > x2
⇔
x<0
8
½
0x2 > −1 toujours vérifié ∀x ∈ R∗−
x<0
dom2 = R∗−
⇔
Donc: domf = dom1 ∪ dom2 = R+ ∪ R∗− = R
¡
¢3
1 + x2
f (x) = log10 p
√
x + 1 + x2
p
√
¡
¢3
= log10 1 + x2 − log10 x + 1 + x2
¯
¯
√
1
= 3 log10 (1 + x2 ) − log10 ¯x + 1 + x2 ¯
2
6.
(a) log10 x = 1(E)
x ∈ dom(E) ⇔ x ∈ R∗+
Donc dom(E) = R∗+
log10 x = 1
⇔ x = 101
⇔ x = 10
S = {10}
(b) log10 x = 3(E)
x ∈ dom(E) ⇔ x ∈ R∗+
Donc dom(E) = R∗+
log10 x = 3
⇔ x = 103
⇔ x = 1000
S = {1000}
(c) log10 x = −4(E)
x ∈ dom(E) ⇔ x ∈ R∗+
Donc dom(E) = R∗+
log10 x = −4
9
⇔ x = 10−4
⇔ x = 10 1000
©
ª
S = 10 1000
(d) log10 (x + 4) + log10 x = 0(E)
x ∈ dom(E) ⇔ x > −4 et x ∈ R∗+
Donc dom(E) = R∗+
log10 (x + 4) + log10 x = 0
⇔ log10 [(x + 4) · x] = 0
⇔ log10 (x2 + 4x) = log 1
⇔ x2 + 4x = 1
⇔ x2 + 4x − 1 = 0
∆ = b2 − 4ac
= 42 − 4 · 1 · (−1)
= 16 + 4
= 20
√
√
∆ = 2 √5
√
√
∆
5
x1 = −b−
= −4−2
= −2 − 5 à rejeter
2a
2
√
√
√
5
∆
x2 = −b+
= −4+2
= −2 + 5
2a
2
√ ª
©
S = −2 + 5
(e) log10 (x + 3) + log10 (x + 5) = log10 15(E)
x ∈ dom(E) ⇔ x > −3 et x > −5
Donc dom(E) = ]−3; +∞[
log10 (x + 3) + log10 (x + 5) = log10 15
⇔ log10 ((x + 3) · (x + 5)) = log10 15
⇔ x2 + 8x + 15 = 15
⇔ x(x + 8) = 0
⇔ x = 0 ou x = −8 à rejeter
S = {0}
(f) log10 (x + 1) = 3 − log10 (1 − 2x)(E)
x ∈ dom(E) ⇔ x > −1 et x < 12
¤
£
Donc dom(E) = −1; 12
log10 (x + 1) = 3 − log10 (1 − 2x)
⇔ log10 (x + 1) + log10 (1 − 2x) = 3
10
⇔ log10 ((x + 1) · (1 − 2x)) = 3
⇔ log10 (−2x2 − x + 1) = 3
⇔ −2x2 − x + 1 = 103
⇔ 2x2 + x − 1001 = 0
∆ = b2 − 4ac = 1 − 4 · 2 · (−1001) = 8009
√
−1− 8009
=
4
√
−1+ 8009
x2 =
4
n
o
√
−1+ 8009
S=
4
x1 =
−22. 623 à rejeter
(g) log10 (1 − x) − log10 (1 + x) = −2(E)
x ∈ dom(E) ⇔ x < 1 et x > −1
Donc dom(E) = ]−1; 1[
log10 (1 − x) − log10 (1 + x) = −2
⇔ log10
⇔
1−x
1+x
(1−x)
(1+x) =
−2
= 10
−2
1
⇔ 1 − x = 100
· (1 + x)
⇔ −100x = 1 + x − 100
⇔ −101x = −99
99
⇔ x = 101
© 99 ª
S = 101
(h) log10 (x + 1) + log10 (x − 1) = log10 3 + 4 log10 2(E)
x ∈ dom(E) ⇔ x > −1 et x > 1
Donc dom(E) = ]1; +∞[
log10 (x + 1) + log10 (x − 1) = log10 3 + 4 log10 2
⇔ log10 ((x + 1) · (x − 1)) = log10 (3 · 24 )
⇔ x2 − 1 = 48
⇔ x2 − 49 = 0
⇔ x = 7 ou x = −7 à rejeter
S = {7}
(i) log10 (x2 + 5x + 6) = log10 (x + 11)(E)
x ∈ dom(E) ⇔ x2 + 5x + 6 > 0 et x > −11
⇔ (x + 2)(x + 3) > 0 et x > −11
⇔ x ∈ ]−∞, −3[ ∪ ]−2; +∞[ et x > −11
11
Donc dom(E) = ]−11; −3[ ∪ ]−2; +∞[
log10 (x2 + 5x + 6) = log10 (x + 11)
⇔ x2 + 5x + 6 = x + 11
⇔ x2 + 5x + 6 − x − 11 = 0
⇔ x2 + 4x − 5 = 0
∆ = 36
= −5
x1 = −4−6
2
=
1
x2 = −4+6
2
⇔ x = −5 ou x = 1
S = {−5, 1}
(j) log10 (1 − 5x) − log10 (x + 1) = −1(E)
x ∈ dom(E) ⇔ x < 15 et x > −1
£
¤
Donc dom(E) = −1; 15
log10 (1 − 5x) + 1 = log10 (x + 1)
⇔ log10 (1 − 5x) · 10 = log10 (x + 1)
⇔ (1 − 5x) · 10 = x + 1
⇔ 10 − 50x − x − 1 = 0
⇔ 51x = 9
3
⇔x=
½ 17
¾
3
S=
17
7.
2
(a) (log10 x) − 3 log10 x − 4 = 0 (E)
dom(E) = R∗+
On pose: y = log10 x
Donc: y 2 − 3y − 4 = 0
⇐⇒ y = −1 ou y = 4
Alors: log10 x = −1 ou log10 x = 4
⇐⇒ x = 10−1 ou x = 104
1
⇐⇒ x =
ou x = 10000
10½
¾
1
D’où: S =
; 10000
10
12
(b) 2 (log10 x)2 − log10 x + 1 = 0 (E)
dom(E) = R∗+
On pose: y = log10 x
Donc:2y 2 − y + 1 = 0
Cette équation n’admet pas de racines réelles.
D’où: S = ∅
(c) (log10 x)2 + log10 x − 12 = 0 (E)
dom(E) = R∗+
On pose: y = log10 x
Donc: y 2 + y − 12 = 0
⇐⇒ y = −4 ou y = 3
Alors: log10 x = −4 ou log10 x = 3
⇐⇒ x = 10−4 ou x = 103
1
⇐⇒ x =
ou x = 1000
10000
½
¾
1
D’où: S =
; 1000
10000
8.
(a) log10 x > 12 (I)
dom(I) = R∗+
Donc: log10 x >
1
2
1
⇐⇒ x > 10 2
√
⇐⇒ x > 10
¤√
£
D’où: S =
10; +∞
(b) 2 log10 x ≤ −3 (I)
dom(I) = R∗+
Donc: 2 log10 x ≤ −3
⇐⇒ log10 x2 ≤ −3
⇐⇒ x2 ≤ 10−3
Ãr
!2
1
2
⇐⇒ x −
≤0
103
13
Ã
!Ã
!
r
10
10
⇐⇒ x −
x+
≤0
103 · 10
103 · 10
# √ #
" √
√ #
10
10
10
;
∩ dom(I) = 0;
D’où: S = −
100 100
100
r
(c) log10 |2x + 1| + log10 |x + 3| < 1 (I)
x ∈ dom(I)
⇐⇒ |2x + 1| > 0 et |x + 3| > 0
⇐⇒ x 6= − 12 et x 6= −3
ª
©
dom(I) = R\ − 12 ; −3
Donc: log10 |2x + 1| + log10 |x + 3| < 1
⇐⇒ log10 [(|2x + 1|) (|x + 3|)] < 1
⇐⇒ (|2x + 1|) (|x + 3|) < 101
⇐⇒ |(2x + 1) (x + 3)| < 10
¯
¯
⇐⇒ ¯2x2 + 6x + x + 3¯ < 10
⇐⇒ ½
−10 < 2x2 + 7x + 3 < 10
−10 < 2x2 + 7x + 3
⇐⇒
2x2 + 7x + 3 < 10
½
2x2 + 7x + 13 > 0 (1)
⇐⇒
2x2 + 7x − 7 < 0 (2)

x∈R


"
#
√
√
−7 − 105 −7 + 105
⇐⇒
;

 x∈
4
4
"
#
√
√
−7 − 105 −7 + 105
;
⇐⇒ x ∈
4
4
"
#
√
√
−7 − 105 −7 + 105
;
S=
4
4
(d) log10 24 + log10 (3 − x) < log10 (x + 1) + log10 (25x − 49) (I)
49
x ∈ dom(I) ⇐⇒ 3 > x et x > −1 et x >
25
·
¸
49
;3
⇐⇒ x ∈
25
·
¸
49
;3
dom(I) =
25
Donc: log10 24 + log10 (3 − x) < log10 (x + 1) + log10 (25x − 49)
⇐⇒ log10 [24 · (3 − x)] < log10 [(x + 1) (25x − 49)]
¡
¢
⇐⇒ log10 (72 − 24x) < log10 25x2 + 25x − 49x − 49
14
¢
¡
⇐⇒ log10 (72 − 24x) < log10 25x2 − 24x − 49
⇔ 72 − 24x < 25x2 − 24x − 49
⇔ 0 < 25x2 − 121
⇔ (5x − 11) (5x + 11) > 0
¸ ·
·
¸
11
11
∪
; +∞
⇐⇒ x ∈ −∞; −
5
5
¸ ·
·¶ ¸
· ·
·
µ¸
11
49
11
11
∪
; +∞ ∩
;3 =
;3
S=
−∞; −
5
5
25
5
¢
¡
(e) log10 3x2 − x − 2 > log10 (6x + 4) (I)
x ∈ dom(I) ⇐⇒ 3x2 − x − 2 > 0 et 6x + 4 > 0
¸
·
2
2
⇐⇒ x ∈ −∞; − ∪ ]1; +∞[ et x > −
3
3
dom(I) = ]1; +∞[
¢
¡
Donc: log10 3x2 − x − 2 > log10 (6x + 4)
¢
¡
⇐⇒ log10 3x2 − x − 2 − log10 (6x + 4) > 0
¢
¡ 2
⇐⇒ log10 3x − x − 2 > log10 (6x + 4)
⇐⇒ 3x2 − x − 2 > 6x + 4
⇐⇒ 3x2 −
¸ 7x − 6 >
·0
2
⇐⇒ x ∈ −∞, − ∪ ]3; +∞[
3
·
¶
µ¸
2
S=
−∞, − ∪ ]3; +∞[ ∩ ]1; +∞[ = ]3; +∞[
3
(f) log10 (x + 2) + log10 (x − 4) < 2 log10 (x − 1) (I)
x ∈ dom(I) ⇐⇒ x + 2 > 0 et x − 4 > 0 et x − 1 > 0
⇐⇒ x > −2 et x > 4 et x > 1 ⇐⇒ x > 4
dom(I) = ]4; +∞[
Donc: log10 (x + 2) + log10 (x − 4) < 2 log10 (x − 1)
2
⇐⇒ log10 [(x + 2) (x − 4)] < log10 (x − 1)
¡ 2
¡ 2
¢
¢
⇐⇒ log10 x − 2x − 8 < log10 x − 2x + 1
⇐⇒ x2 − 2x − 8 < x2 − 2x + 1
⇐⇒ 0x < 9 vrai ∀x ∈ dom(I)
D’où: S = ]4; +∞[
Saisie et mise en page du corrigé :
Exercices 1-3: Alain KLEIN, IIe C2 (2007-08)
Exercices 4-5: Ailin ZHANG, IIe B2 (2007-08)
Exercice 6: Bob WEBER, IIe B2 (2007-08)
Exercices 7-8: Bob HEYMANS, IIe B2 (2007-08)
15