Logarithmes décimaux

Exercices sur le logarithme

décimal

1. Soient a et b ∈ R∗+. Simplifier: (a) log 0, 1 · Ã a2 r b2 a !3 a b3 (b) log µ 10a3_b−2 a√a2_b3 ¶3µ a−4_b3 100√4b2_a ¶−2 (c) log 0, 001³√3a4_b−2´3 √ b3√4 a3 (d) log³ 10−3a4 3√b 0,01a2√_a3_b2 ´ 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; log1₄; log 0, 2

4. Si log b = a avec b ∈ R∗

+, alors déterminer:

log 10b; log₁₀₀b ; log1_b; log√b; log b5_{; 2 log 3b + log}√5

b − log 9 5. Déterminer domf et simplifier f (x) si possible:

(a) f (x) = log(4 − 3x) (b) f (x) = log(4 − x2₎ (c) f (x) = log(2x − 3) 3 2 − x (d) f (x) = log4x − 1 x − 3 (e) f (x) = log |5x − 1|

(f) f (x) = logx 3_{− 6x}2_{+ 11x − 6} x2_{+ 3x + 2} (g) f (x) = log (1 + x 2₎3 p x +√1 + x2

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 x6 −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)}

Corrigé

1. (a) log100.1 Ã a2 r b2 a !3 a b3 = log₁₀0.1 + log₁₀ Ã a2 r b2 a !3 + log₁₀ a b3 = −1 + log10a6+ log10 r b2 a 3 + log₁₀ a b3 = −1 + 6 log10a + 3 2log10 b2 a + log10a − log10b 3 = −1 + 6 log10a + 3 log10b − 3

2log10a + log10a − 3 log10b = −1 +11₂ log₁₀a (b) log₁₀ µ 10a3_b−2 a√a2_b3 ¶3µ a−4_b3 100√4b2_a ¶−2 = 3 log₁₀10a 3_b−2 a√a2_b3 − 2 log10 a−4_b3 100√4b2_a

= 3 log₁₀10+3 log₁₀a3_{+3 log}

10b−2−3 log10a− 3 2log10a 2₋3 2log10b 3₋

2 log₁₀a−4_{− 2 log10}b3_{+ 2 log}

10100 +

2 4log10b

2₊2

4log10a = 3 + 9 log_{10a − 6 log10b − 3 log10a − 3 log10a −}9

2log10b + 8 log10a − 6 log10b + 4 + log10b + 1 2log10a = 7 + 23 2 log10a − 31 2 log10b (c) log₁₀ 0.001³√3 a4_b−2´3 √ b3√4 a3 = log₁₀0.001 + 3 log₁₀√3 a4_b−2_{− log} 10 √ b3_{− log} 10 4 √ a3 = −3 + log10a4+ 3 log10b−2− 1 2log10b 3₋1 4log10a 3

= −3 + 4 log10a − 6 log10b − 3 2log10b − 3 4log10a = −3 +134 log10a − 15 2 log10b (d) log₁₀ Ã 10−3_a4√3 b 0.01a2√_a3_b2 ! = log₁₀10−3_{+ log} 10a4+ log10 3 √

b − log100.01 − log10a2− log10

√ a3_b2 = −3 + 4 log10a + 1 3log10b + 2 − 2 log10a − 3 2log10a − log10b = −1 +1₂log_{10a −}2 3log10b 2.

(a) log₁₀2 + log₁₀5 = log_{10(2 · 5)} = log₁₀10 = 1

(b) 2 log₁₀5 + log_{1012 − log10}3 = log₁₀52_{+ log} 1012 − log103 = log10 µ 25 · 12 3 ¶ = log₁₀ µ 300 3 ¶ = log₁₀100 = 2 3. (a) log₁₀4 = log1022 = 2 log₁₀2 = 2a

(b) log₁₀16 = log₁₀24 = 4 log₁₀2 = 4a (c) log₁₀40 = log10(4 · 10) = log₁₀4 + log₁₀10 = 2a + 1 (d) log₁₀1 4 = log_{101 − log10}4 = 0 − 2a = −2a (e) log₁₀0.2 = log_{10(0.1 · 2)} = log₁₀0.1 + log₁₀2 = −1 + a 4.

(a) log1010b log1010 + log10b = 1 + a

(b) log10 b 100 = log_{10b − log10}100 = a − 2 (c) log₁₀1 b = log101 − log10b = 0 − a = −a

(d) log₁₀√b = 1₂log₁₀b = 1₂a (e) log₁₀b5 = 5 log₁₀b = 5a (f) 2 log103b + log10 5 √ b − log109

= 2 (log103 + log10b) +15log10b − log1032

= 2 log₁₀3 + 2 log₁₀b +1 5log10b − 2 log103 = 2a + 1 5a = 11 5 5. (a) f (x) = log10(4 − 3x) x ∈ domf ⇔ 4 − 3x > 0 ⇔ x < 43 ⇔ x ∈ ¤ −∞;43 £ domf =¤_−∞;4 3 £

f (x) n’est pas simplifiable!

(b) f (x) = log_{10(4 − x}2₎

x ∈ domf ⇔ 4 − x2> 0 ⇔ (2 − x)(2 + x) > 0 ⇔ x ∈ ]−2; 2[ domf = ]−2; 2[

f (x) = log_{10[(2 − x)(2 + x)]} = log_{10|2 − x| + log10|2 + x|}

(c) f (x) = log₁₀(2x − 3) 2 2 − x x ∈ domf ⇔ (2x − 3) 2 2 − x > 0 et 2 − x 6= 0 ⇔ x ∈ ]−∞, 2[ \ ½ 3 2 ¾ domf = ]−∞, 2[ \ ½ 3 2 ¾ f (x) = log10(2x − 3) 2 2 − x = log_{10(2x − 3)}2_{− log} 10|2 − x| = 2 log_{10|2x − 3| − log10|2 − x|} (d) f (x) = log₁₀4x − 1 x − 3 x ∈ domf ⇔ 4x − 1 x − 3 > 0 et x − 3 6= 0 ⇔ x ∈ ¸ −∞;1₄ · ∪ ]3; +∞[ domf = ¸ −∞;1₄ · ∪ ]3; +∞[ f (x) = log₁₀4x − 1 x − 3 = log_{10|4x − 1| − log10|x − 3|} (e) f (x) = log_{10|5x − 1|} x ∈ domf ⇔ |5x − 1| > 0 ⇔ x ∈ R\ ½ 1 5 ¾ domf = R\ ½ 1 5 ¾

(f) f (x) = log₁₀x 3_{− 6x}2_{+ 11x − 6} x2_{+ 3x + 2} x ∈ domf ⇔ x 3_{− 6x}2_{+ 11x − 6} x2_{+ 3x + 2} > 0 et x 2_{+ 3x + 2 6= 0} ⇔ (x − 1)(x − 2)(x − 3)_{(x + 1)(x + 2)} > 0 et (x + 2)(x + 1) 6= 0 ⇔ x ∈ ]−2; −1[ ∪ ]1; 2[ ∪ ]3; +∞[ domf = ]−2; −1[ ∪ ]1; 2[ ∪ ]3; +∞[ f (x) = log₁₀x 3_{− 6x}2_{+ 11x − 6} x2_{+ 3x + 2} = log₁₀(x − 1)(x − 2)(x − 3) (x + 2)(x + 1) = log_{10|(x − 1)(x − 2)(x − 3)| − log10|(x + 2)(x + 1)|}

= log_{10|x − 1|+log10|x − 2|+log10|x − 3|−log10|x + 2|−log10|x + 1|}

(g) f (x) = log₁₀ ¡ 1 + x2¢3 p x +√1 + x2 x ∈ domf ⇔ ¡ 1 + x2¢3 p x +√1 + x2 > 0 et x + √ 1 + x2_{> 0 et 1 + x}2 ≥ 0 ⇔ x +√1 + x2_{> 0} ⇔√1 + x2_{> −x} 1er _{cas: x ∈ R} +: √ 1 + x2_{> −x toujours vérifié ∀x ∈ R} + ⇔ x ∈ R+ dom1= R+ 2e_{cas: x ∈ R}∗₋: √ 1 + x2_{> −x} ⇔ ½ 1 + x2_{> x}2 x < 0

⇔ ½

0x2_{> −1 toujours vérifié ∀x ∈ R}∗₋ x < 0

dom2= R∗₋

Donc: domf = dom1∪ dom2= R+∪ R∗₋ = R

f (x) = log₁₀ ¡ 1 + x2¢3 p x +√1 + x2 = log10 ¡ 1 + x2¢3_{− log}10 p x +√1 + x2 = 3 log₁₀(1 + x2_{) −}1 2log10 ¯ ¯x +√1 + x2¯¯ 6.

(a) log₁₀x = 1(E)

x ∈ dom(E) ⇔ x ∈ R∗+ Donc dom(E) = R∗ + log₁₀x = 1 ⇔ x = 101 ⇔ x = 10 S = {10} (b) log10x = 3(E) x ∈ dom(E) ⇔ x ∈ R∗ + Donc dom(E) = R∗ + log₁₀x = 3 ⇔ x = 103 ⇔ x = 1000 S = {1000} (c) log10x = −4(E) x ∈ dom(E) ⇔ x ∈ R∗ + Donc dom(E) = R∗ + log_{10x = −4}

⇔ x = 10−4

⇔ x = 10 0001

S =©10 0001

ª

(d) log10(x + 4) + log10x = 0(E)

x ∈ dom(E) ⇔ x > −4 et x ∈ R∗ + Donc dom(E) = R∗ + log₁₀(x + 4) + log₁₀x = 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 x1=−b− √ ∆ 2a =−4−2 √ 5 2 = −2 − √ 5 à rejeter x2=−b+ √ ∆ 2a =−4+2 √ 5 2 = −2 + √ 5 S = ©−2 +√5ª

(e) log₁₀(x + 3) + log₁₀(x + 5) = log₁₀15(E) x ∈ dom(E) ⇔ x > −3 et x > −5 Donc dom(E) = ]−3; +∞[

log₁₀(x + 3) + log₁₀(x + 5) = log₁₀15 ⇔ log10((x + 3) · (x + 5)) = log1015

⇔ x2+ 8x + 15 = 15 ⇔ x(x + 8) = 0

⇔ x = 0 ou x = −8 à rejeter S = {0}

(f) log_{10(x + 1) = 3 − log10(1 − 2x)(E)} x ∈ dom(E) ⇔ x > −1 et x < 12

Donc dom(E) =¤_−1;1₂£

log_{10(x + 1) = 3 − log10(1 − 2x)} ⇔ log10(x + 1) + log10(1 − 2x) = 3

⇔ 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} x1=−1− √ 8009 4 = −22. 623 à rejeter x2=−1+ √ 8009 4 S =n−1+√48009 o

(g) log_{10(1 − x) − log10(1 + x) = −2(E)} x ∈ dom(E) ⇔ x < 1 et x > −1 Donc dom(E) = ]−1; 1[ log_{10(1 − x) − log10(1 + x) = −2} ⇔ log10 (1−x) (1+x) = −2 ⇔ 11+x−x = 10−2 ⇔ 1 − x = 1001 · (1 + x) ⇔ −100x = 1 + x − 100 ⇔ −101x = −99 ⇔ x = 99 101 S =©10199 ª

(h) log₁₀(x + 1) + log_{10(x − 1) = log10}3 + 4 log₁₀2(E) x ∈ dom(E) ⇔ x > −1 et x > 1

Donc dom(E) = ]1; +∞[

log10(x + 1) + log10(x − 1) = log103 + 4 log102

⇔ 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

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 x1=−4−6₂ = −5 x2=−4+6₂ = 1 ⇔ x = −5 ou x = 1 S = {−5, 1}

(j) log_{10(1 − 5x) − log10(x + 1) = −1(E)} x ∈ dom(E) ⇔ x < 15 et x > −1 Donc dom(E) =¤_−1;1₅£ log_{10(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 ⇔ x = ₁₇3 S = ½ 3 17 ¾ 7.

(a) (log₁₀x)2_{− 3 log10x − 4 = 0 (E)} dom(E) = R∗

+

On pose: y = log₁₀x Donc: y2_{− 3y − 4 = 0}

⇐⇒ y = −1 ou y = 4

Alors: log_{10x = −1 ou log10}x = 4 ⇐⇒ x = 10−1 _{ou x = 10}4 ⇐⇒ x = 1 10 ou x = 10000 D’où: S = ½ 1 10; 10000 ¾

(b) 2 (log₁₀x)2_{− log10}x + 1 = 0 (E) dom(E) = R∗

+

On pose: y = log₁₀x Donc:2y2_{− y + 1 = 0}

Cette équation n’admet pas de racines réelles. D’où: S = ∅

(c) (log₁₀x)2+ log_{10x − 12 = 0 (E)} dom(E) = R∗

+

On pose: y = log₁₀x Donc: y2_{+ y − 12 = 0}

⇐⇒ y = −4 ou y = 3

Alors: log_{10x = −4 ou log10}x = 3 ⇐⇒ x = 10−4 ou x = 103 ⇐⇒ x = ₁₀₀₀₀1 ou x = 1000 D’où: S = ½ 1 10000; 1000 ¾ 8.

(a) log₁₀x > 1₂ (I) dom(I) = R∗ + Donc: log₁₀x > 1₂ ⇐⇒ x > 1012 ⇐⇒ x >√10 D’où: S =¤√10; +∞£ (b) 2 log_{10x ≤ −3 (I)} dom(I) = R∗ + Donc: 2 log_{10x ≤ −3} ⇐⇒ log10x2≤ −3 ⇐⇒ x2≤ 10−3 ⇐⇒ x2₋ Ãr 1 103 !2 ≤ 0

⇐⇒ Ã x − r 10 103_{· 10} ! Ã x + r 10 103_{· 10} ! ≤ 0 D’où: S = " − √ 10 100; √ 10 100 # ∩ dom(I) = # 0; √ 10 100 #

(c) log_{10|2x + 1| + log10|x + 3| < 1 (I)} x ∈ dom(I) ⇐⇒ |2x + 1| > 0 et |x + 3| > 0 ⇐⇒ x 6= −12 et x 6= −3 dom(I) = R\©−1 2; −3 ª

Donc: log_{10|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 ⇐⇒ −10 < 2x2_{+ 7x + 3 < 10} ⇐⇒ ½ −10 < 2x2_{+ 7x + 3} 2x2_{+ 7x + 3 < 10} ⇐⇒ ½ 2x2_{+ 7x + 13 > 0 (1)} 2x2_{+ 7x − 7 < 0 (2)} ⇐⇒      x ∈ R x ∈ # −7 −√105 4 ; −7 +√105 4 " ⇐⇒ x ∈ # −7 −√105 4 ; −7 +√105 4 " S = # −7 −√105 4 ; −7 +√105 4 "

(d) log₁₀24 + log_{10(3 − x) < log10}(x + 1) + log_{10(25x − 49) (I)} x ∈ dom(I) ⇐⇒ 3 > x et x > −1 et x > 49₂₅ ⇐⇒ x ∈ ¸ 49 25; 3 · dom(I) = ¸ 49 25; 3 ·

Donc: log₁₀24 + log_{10(3 − x) < log10}(x + 1) + log_{10(25x − 49)} ⇐⇒ log10[24 · (3 − x)] < log10[(x + 1) (25x − 49)]

⇐⇒ log10(72 − 24x) < log10

¡

⇐⇒ log10(72 − 24x) < log10 ¡ 25x2_{− 24x − 49}¢ ⇔ 72 − 24x < 25x2− 24x − 49 ⇔ 0 < 25x2_{− 121} ⇔ (5x − 11) (5x + 11) > 0 ⇐⇒ x ∈ ¸ −∞; −11₅ ¸ ∪ · 11 5; +∞ · S = µ¸ −∞; −11 5 ¸ ∪ · 11 5; +∞ ·¶ ∩ ¸ 49 25; 3 · = · 11 5; 3 ·

(e) log₁₀¡3x2_{− x − 2}¢_{> log}

10(6x + 4) (I) x ∈ dom(I) ⇐⇒ 3x2− x − 2 > 0 et 6x + 4 > 0 ⇐⇒ x ∈ ¸ −∞; −2₃ · ∪ ]1; +∞[ et x > −2₃ dom(I) = ]1; +∞[

Donc: log₁₀¡3x2_{− x − 2}¢_{> log}

10(6x + 4) ⇐⇒ log10 ¡ 3x2_{− x − 2}¢_{− log} 10(6x + 4) > 0 ⇐⇒ log10 ¡ 3x2_{− x − 2}¢_{> log} 10(6x + 4) ⇐⇒ 3x2_{− x − 2 > 6x + 4} ⇐⇒ 3x2_{− 7x − 6 > 0} ⇐⇒ x ∈ ¸ −∞, −2₃ · ∪ ]3; +∞[ S = µ¸ −∞, −2₃ · ∪ ]3; +∞[ ¶ ∩ ]1; +∞[ = ]3; +∞[ (f) log₁₀(x + 2) + log_{10(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) + log_{10(x − 4) < 2 log10(x − 1)} ⇐⇒ log10[(x + 2) (x − 4)] < log10(x − 1) 2 ⇐⇒ log10 ¡ x2_{− 2x − 8}¢_{< log} 10 ¡ x2_{− 2x + 1}¢ ⇐⇒ x2_{− 2x − 8 < x}2_{− 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)}