داﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ اﻟﺜﺎﻧﻴﺔ ﺳﻠﻚ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ -Іداﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ اﻟﻨﻴﺒﻴﺮي -1ﺗﺬآﻴﺮ -ﻧﻌﻠﻢ أن آﻞ داﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ ﻣﺠﺎل Iﺗﻘﺒﻞ دوال أﺻﻠﻴﺔ ﻋﻠﻰ I
r x r +1 اﻟﺪاﻟﺔ x →xﺗﻘﺒﻞ دوال أﺻﻠﻴﺔ ﻋﻠﻰ [∞ ]0; +هﻲ + k r +1
-ﻧﻌﻠﻢ أن ﻟﻜﻞ rﻣﻦ }− {−1
→x
ﺣﻴﺚ kﻋﺪد ﺣﻘﻴﻘﻲ ﺛﺎﺑﺖ 1 → xاﻟﻤﺘﺼﻠﺔ ﻋﻠﻰ [∞ ]0; +وﻣﻨﻪ ﺗﻘﺒﻞ دوال أﺻﻠﻴﺔ * -ﻓﻲ اﻟﺤﺎﻟﺔ اﻟﺘﻲ ﺗﻜﻮن r=-1ﻧﺤﺼﻞ ﻋﻠﻰ اﻟﺪاﻟﺔ x 1 وﺑﺎﻟﺘﺎﻟﻲ اﻟﺪاﻟﺔ → xﺗﻘﺒﻞ داﻟﺔ أﺻﻠﻴﺔ وﺣﻴﺪة ﺗﻨﻌﺪم ﻓﻲ .1 x -2ﺗﻌﺮﻳﻒ 1 اﻟﺪاﻟﺔ اﻷﺻﻠﻴﺔ ﻟﺪاﻟﺔ → xﻋﻠﻰ [ ∞ ]0; +اﻟﺘﻲ ﺗﻨﻌﺪم ﻓﻲ اﻟﻨﻘﻄﺔ 1ﺗﺴﻤﻰ داﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ اﻟﻨﻴﺒﻴﺮي x و ﻳﺮﻣﺰ ﻟﻬﺎ ﺑﺎﻟﺮﻣﺰ lnأو Log 0 x 1 ) f ' ( x ) = ⇔ f ( x ) = ln( x x f (1) = 0 -3ﺧﺎﺻﻴﺎت أ -ﺧﺎﺻﻴﺎت * -ﻣﺠﻤﻮﻋﺔ ﺗﻌﺮﻳﻒ اﻟﺪاﻟﺔ lnهﻲ * -اﻟﺪاﻟﺔ lnﻣﺘﺼﻠﺔ ﻋﻠﻰ
[∞]0; +
ln(1)=0
[∞]0; +
* -اﻟﺪاﻟﺔ lnﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ * -اﻟﺪاﻟﺔ lnﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ ﻋﻠﻰ
[∞]0; +
[∞]0; +
1 و x
[ ∞∀ x ∈ ]0; +
= ) ln'( x
ﻧﺘﺎﺋﺞ ﻟﻜﻞ ﻋﺪدﻳﻦ ﺣﻘﻴﻘﻴﻴﻦ ﻣﻮﺟﺒﻴﻦ ﻗﻄﻌﺎ xو y
ln x = ln y ⇔ x = y
y ﻣﻼﺣﻈﺔ
ln y ⇔ x
ln x
ln x = 0 ⇔ x = 1
ln x 0 ⇔ x 1 ln x ≺ 0 ⇔ 0 ≺ x ≺ 1 ﺗﻤﺮﻳﻦ -1ﺣﺪد ﻣﺠﻤﻮﻋﺔ ﺗﻌﺮﻳﻒ اﻟﺪاﻟﺘﻴﻦ ) f : x → ln ( x − 1) + ln ( 4 − x -2ﺣﻞ ﻓﻲ -3ﺣﻞ ﻓﻲ
(
) ln ( x − x − 2 ) ≺ 0
اﻟﻤﻌﺎدﻟﺘﻴﻦ ln x 2 + 2x = 0 اﻟﻤﺘﺮاﺟﺤﺘﻴﻦ
2
)
)
(
g : x → ln x 2 − 3x
(
) ln x 2 − 3 = ln ( 2x
)
(
) ln x 2 − 2x ≤ ln ( x
ب -ﺧﺎﺻﻴﺔ أﺳﺎﺳﻴﺔ ﻧﺸﺎط ﻟﻴﻜﻦ aو bﻋﺪدﻳﻦ ﺣﻘﻴﻘﻴﻴﻦ ﻣﻮﺟﺒﻴﻦ ﻗﻄﻌﺎ و Fداﻟﺔ ﻋﺪدﻳﺔ ﻣﻌﺮﻓﺔ ﻋﻠﻰ [∞ ]0; +ﺑـ )F ( x ) = ln(ax
1 1 [∞ ∀x ∈ ]0; +و اﺳﺘﻨﺘﺞ ان Fداﻟﺔ أﺻﻠﻴﺔ ﻟﺪاﻟﺔ -1ﺑﻴﻦ أن = ) F ' ( x x x -2ﺑﻴﻦ أن ∀x ∈ ]0; +∞[ F ( x ) = ln ( ax ) = ln a + ln xﺛﻢ اﺳﺘﻨﺘﺞ ln ( ab ) = ln a + ln b
→ xﻋﻠﻰ
[∞]0; +
اﻟﺠﻮاب ﻟﺪﻳﻨﺎ-1
u ( x ) = ax ﺣﻴﺚF ( x ) = ln u ( x )
F ' ( x ) = u ' ( x ) × ( ln ) ' ( u ( x ) ) = a ⋅
∀x ∈ ]0; +∞[
]0; +∞[
ﻋﻠﻰx →
1 1 = ax x
1 داﻟﺔ أﺻﻠﻴﺔ ﻟﺪاﻟﺔF و ﻣﻨﻪ x
1 داﻟﺘﺎن أﺻﻠﻴﺘﺎن ﻟﺪاﻟﺔx → ln x وF ﻟﺪﻳﻨﺎ-2 x ∀ x ∈ ]0; +∞ [ F ( x ) = k + ln x اذن
]0; +∞[
ﻋﻠﻰx →
k = ln a
F (1) = k
و ﻣﻨﻪ
∀ x ∈ ]0; +∞ [
F (1) = ln ( a )
و
ﻟﺪﻳﻨﺎ
F ( x ) = ln ( ax ) = ln a + ln x
إذن
ln ( ab ) = ln a + ln b ﻧﺤﺼﻞ ﻋﻠﻰx = b ﺑﻮﺿﻊ
ﺧﺎﺻﻴﺔ أﺳﺎﺳﻴﺔ ∀ ( a ; b ) ∈ ( ]0; +∞ [ )
2
ln ( ab ) = ln a + ln b
ﺧﺎﺻﻴﺎت-ج
∀x ∈ ]0; +∞[
ln
∀ ( x; y ) ∈ ]0; +∞[
2
1 = − ln x x x ln = ln x − ln y y
∀ ( x1 ; x2 ;....; xn ) ∈ ]0; +∞[ ∀x ∈ ]0; +∞[
∀r ∈
n
*
ln ( x1 × x2 × .......... × xn ) = ln x1 + ln x2 + .... + ln xn ln x r = r ln x اﻟﺒﺮهﺎن
1 1 1 ln x × = ln1 ⇔ ln x + ln = 0 ⇔ ln = − ln x x x x ln x r = ln ( x × x × ....... × x ) = ln x + ln x + ....... + ln x = r ln x ﻓﺎنr ∈ r
r
facteurs
ln x r = ln x − n = ln
1 x
n
ln x
=
p ln x q
اذن
ln y =
إذا آﺎن
p ln x q
p q
termes
* −
= − ln x n = − n ln x = r ln x وﻣﻨﻪr = − n ﻓﺈﻧﻨﺎ ﻧﻀﻊr ∈ y=x
p q
*
⇔ x p = y q ﻧﻌﻠﻢ أن
q∈
أيp ln x = q ln y و ﺑﺎﻟﺘﺎﻟﻲ
*
ln x
p∈ p
*
/
إذا آﺎن
p = r إذا آﺎن q
= ln y q و ﻣﻨﻪ ln x r = r ln x أي
1 ln x ﺣﺎﻟﺔ ﺧﺎﺻﺔ 2 ﻣﺘﺴﺎوﻳﺘﻴﻦ ﻓﻲ اﻟﺤﺎﻟﺘﻴﻦ اﻟﺘﺎﻟﻴﺘﻴﻦg وf هﻞ اﻟﺪاﻟﺘﺎن ﺗﻤﺮﻳﻦ ∀x ∈ ]0; +∞[
f ( x ) = ln ( x − 1)
g ( x ) = 2ln x − 1
2
f ( x ) = ln x ( x − 1)
g ( x ) = ln x + ln ( x − 1)
0, 7
(a (b
2 + 1 + ln 2 −1 ( أﺣﺴﺐ1 ﺗﻤﺮﻳﻦ 2 ln 3 1,1 ادا ﻋﻠﻤﺖ أنln وln 6 ( أﺣﺴﺐ ﻗﻴﻤﺔ ﻣﻘﺮﺑﺔ ﻟـ2 9 ln
ln 2
ln x =
-4دراﺳﺔ داﻟﺔ ln (aداﻟﺔ lnﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ ﻋﻠﻰ [∞]0; +
∞lim ln x = +
(bﻣﺒﺮهﻨﺔ) 1ﻧﻘﺒﻞ(
∞x →+
∞lim ln x = −
ﻣﺒﺮهﻨﺔ2 اﻟﺒﺮهﺎن
1 ﻧﻀﻊ t
x →0 +
= x
1 ∞lim+ ln x = lim ln = lim − ln t = − t →+∞ t ∞t →+ x→0
(cاﻟﻌﺪدe ﻟﺪﻳﻨﺎ اﻟﺪاﻟﺔ lnﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ ﻋﻠﻰ [∞ ]0; +وﻣﺘﺼﻠﺔ و
= ) [∞ ln ( ]0; +و ﻣﻨﻪ اﻟﻤﻌﺎدﻟﺔ ln x = 1ﺗﻘﺒﻞ ﺣﻼ
وﺣﻴﺪا ﻓﻲ [∞ ]0; +وﻳﺮﻣﺰ ﻟﻪ ﺑﺎﻟﺤﺮف eادن ln e = 1
ﻧﻘﺒﻞ أن eﻟﻴﺲ ﻋﺪدا ﺟﺬرﻳﺎ و ﻗﻴﻤﺘﻪ اﻟﻤﻘﺮﺑﺔ هﻲ (dﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ ln e ∞+ ∞+ 1
2, 71828
e 0
1
x
f
0
∞− (eاﻟﻔﺮوع اﻟﻼﻧﻬﺎﺋﻴﺔ
ﺑﻤﺎ أن ∞ lim ln x = −ﻓﺎن ﻣﺤﻮر اﻻراﺗﻴﺐ ﻣﻘﺎرب ﻟﻠﻤﻨﺤﻨﻰ اﻟﻤﻤﺜﻞ اﻟﺪاﻟﺔ ln x →0+
ln x =0 ﻣﺒﺮهﻨﺔ x→+∞ x اذن اﻟﻤﻨﺤﻨﻰ اﻟﻤﻤﺜﻞ ﻟﺪاﻟﺔ lnﻳﻘﺒﻞ ﻓﺮﻋﺎ ﺷﻠﺠﻤﻴﺎ ﻓﻲ اﺗﺠﺎﻩ ﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ 1 [∞ ∀x ∈ ]0; +اذن ﻣﻨﺤﻨﻰ اﻟﺪاﻟﺔ lnﻣﻘﻌﺮ (fدراﺳﺔ اﻟﺘﻘﻌﺮ ( ln ) '' ( x ) = − 2 x (gاﻟﺘﻤﺜﻴﻞ اﻟﻤﺒﻴﺎﻧﻲ lim
ﻣﻨﺤﻨﻰ اﻟﺪاﻟﺔ ln
(hﻧﻬﺎﻳﺎت هﺎﻣﺔ أﺧﺮى ﺧﺎﺻﻴﺔ ln x =1 x→1 x − 1
lim x ln x = 0
lim
x →0 +
ﺣﺪد
ﺗﻤﺮﻳﻦ
=1
lim x − ln x
∞x→+
) ln (1 + x
lim
x x−2 lim x ln ∞x→+ x x →0
=0
ln x n
x
)
lim
∞x→+
lim+ x n ln x = 0ﺣﻴﺚ
(
*
x→0
lim− x ln x 2 − x
x →0
– 5ﻣﺸﺘﻘﺔ اﻟﺪاﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻤﻴﺔ أ -ﻣﺒﺮهﻨﺔ uداﻟﺔ ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ ﻣﺠﺎل Iو ﻻ ﺗﻨﻌﺪم ﻋﻠﻰ هﺬا اﻟﻤﺠﺎل I ) u '( x ∀x ∈ I = ' ) ln u ( x )u (x
(
)
uﻻ ﺗﻨﻌﺪم ﻋﻠﻰ Iو ﻣﻨﻪ uإﻣﺎ ﻣﻮﺟﺒﺔ ﻗﻄﻌﺎ ﻋﻠﻰ Iأو ﺳﺎﻟﺒﺔ ﻗﻄﻌﺎ ﻋﻠﻰ I اﻟﺒﺮهﺎن )u '( x وﻣﻨﻪ اذا آﺎﻧﺖ uﻣﻮﺟﺒﺔ ﻗﻄﻌﺎ ﻋﻠﻰ Iﻓﺎن ) f ( x ) = ln u ( x = ) f ' ( x ) = u ' ( x ) ln' u ( x )u ( x اذا آﺎﻧﺖ uﺳﺎﻟﺒﺔ ﻗﻄﻌﺎ ﻋﻠﻰ Iﻓﺎن )) f ( x ) = ln(−u ( x )u '( x )u ( x
ﺗﻤﺮﻳﻦ
=
) −u ' ( x ) −u ( x
= )) f ' ( x ) = − u ' ( x ) ln'( − u ( x
∀x ∈ I
وﻣﻨﻪ ∀x ∈ I
ﺣﺪد ﻣﺠﻤﻮﻋﺔ ﺗﻌﺮﻳﻒ اﻟﺪاﻟﺔ fو أﺣﺴﺐ ﻣﺸﺘﻘﺘﻬﺎ ﻓﻲ اﻟﺤﺎﻟﺘﻴﻦ اﻟﺘﺎﻟﻴﺘﻴﻦ f ( x ) = ln ( x 2 + 2 x ) (b f ( x ) = ln x 2 − 4 (a
ب -ﺗﻌﺮﻳﻒ uداﻟﺔ ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ ﻣﺠﺎل Iو ﻻ ﺗﻨﻌﺪم ﻋﻠﻰ اﻟﻤﺠﺎل I 'u ﺗﺴﻤﻰ اﻟﻤﺸﺘﻘﺔ اﻟﻠﻮﻏﺎرﻳﺘﻤﻴﺔ ﻟﻠﺪاﻟﺔ uﻋﻠﻰ اﻟﻤﺠﺎل I اﻟﺪاﻟﺔ u ج -ﻧﺘﻴﺠﺔ uداﻟﺔ ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ ﻣﺠﺎل Iو ﻻ ﺗﻨﻌﺪم ﻋﻠﻰ اﻟﻤﺠﺎل I )u '( x → xﻋﻠﻰ Iهﻲ اﻟﺪوال x → ln u ( x ) + cﺣﻴﺚ cﻋﺪد ﺛﺎﺑﺖ اﻟﺪوال اﻷﺻﻠﻴﺔ ﻟﺪاﻟﺔ )u ( x أوﺟﺪ داﻟﺔ أﺻﻠﻴﺔ ﻟﺪاﻟﺔ fﻋﻠﻰ اﻟﻤﺠﺎل Iﻓﻲ اﻟﺤﺎﻻت اﻟﺘﺎﻟﻴﺔ
ﺗﻤﺮﻳﻦ1
x −1 f ( x ) = 2 x − 2x [∞ I = ]2; +
) f ( x ) = tan ( x −π π I = 2 ;2
ﺗﻤﺮﻳﻦ 2أﺣﺴﺐ اﻟﺪاﻟﺔ اﻟﻤﺸﺘﻘﺔ ﻟﺪاﻟﺔ fﻋﻠﻰ [∞ ]−1; +ﺣﻴﺚ
x −1 = ) f ( x x +1 [∞ I = ]−1; +
x3 + 1
( x + 2 )2
= )f ( x
-IIداﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ ﻟﻸﺳﺎس a -1ﺗﻌﺮﻳﻒ aﻋﺪد ﺣﻘﻴﻘﻲ ﻣﻮﺟﺐ ﻗﻄﻌﺎ و ﻣﺨﺎﻟﻒ ﻟﻠﻌﺪد 1 ln x → xاﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ [∞ ]0; +ﺗﺴﻤﻰ داﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ ﻟﻸﺳﺎس aوﻧﺮﻣﺰ ﻟﻬﺎ ﺑﺎﻟﺮﻣﺰ اﻟﺪاﻟﺔ ln a
ln x ln a
= ) Log a ( x
Log a
[∞∀x ∈ ]0; +
ﻣﻼﺣﻈﺎت * -داﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ اﻟﻨﻴﺒﻴﺮي هﻲ داﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ ﻟﻸﺳﺎس e
*-
) (
Log a a r = r
Log a ( a ) = 1
∈ ∀r
}− {1
ln x = ln x ln e *+
∈ ∀a
= ) Loge ( x
[∞∀x ∈ ]0; +
∈n
-2ﺧﺎﺻﻴﺎت ﺑﻤﺎ أن ﻟﻜﻞ xﻣﻦ [∞]0; +
Log a ( x ) = k ln xﺣﻴﺚ kﻋﺪد ﺣﻘﻴﻘﻲ ﺛﺎﺑﺖ ﻓﺎن اﻟﺪاﻟﺔ Log a
ﺗﺤﻘﻖ ﺟﻤﻴﻊ اﻟﺨﺎﺻﻴﺎت اﻟﺘﻲ ﺗﺤﻘﻘﻬﺎ اﻟﺪاﻟﺔ ln ) Log a ( xy ) = Log a ( x ) + Log a ( y
∈ ∀r
2
) [∞∀ ( x; y ) ∈ ( ]0; +
) (
x ) Log a = Log a ( x ) − Log a ( y ) ; Log a x r = rLog a ( x y -3دراﺳﺔ داﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ ﻟﻸﺳﺎس a 1 [∞∀x ∈ ]0; + = ) Log a ' ( x x ln a * -اذا آﺎن 0 ≺ a ≺ 1ﻓﺎن ln a ≺ 0و ﻣﻨﻪ ∀x ∈ ]0; +∞[ Log a ' ≺ 0اذن Log aﺗﻨﺎﻗﺼﻴﺔ ﻗﻄﻌﺎ ﻋﻠﻰ [∞]0; + ∞lim Log a x = +
x →0 +
* -اذا آﺎن 1
aﻓﺎن 0
ln aو ﻣﻨﻪ 0
' Log a
∞lim Log a x = −
x →0 +
∞lim Log a x = −
∞x →+
[∞ ∀x ∈ ]0; +ادن Log aﺗﺰاﻳﺪﻳﺔ
ﻗﻄﻌﺎ ﻋﻠﻰ [∞]0; +
∞lim Log a x = +
∞x →+
-4ﺣﺎﻟﺔ ﺧﺎﺻﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ اﻟﻌﺸﺮي ﺗﻌﺮﻳﻒ اﻟﺪاﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻤﻴﺔ اﻟﺘﻲ أﺳﺎﺳﻬﺎ 10ﺗﺴﻤﻰ داﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ اﻟﻌﺸﺮي و ﻳﺮﻣﺰ ﻟﻬﺎ ﺑـ log ln x [∞∀x ∈ ]0; + = log x = Log10 x ln10 ﻣﻼﺣﻈﺎت 1 [∞∀x ∈ ]0; + = Mﻓﺎﻧﻨﺎ ﻧﺤﺼﻞ ﻋﻠﻰ log x = M ln x * -اذا وﺿﻌﻨﺎ ) 0, 434 ln10 ∈ ∀m log10m = m *- log 0, 01 ﺗﻤﺮﻳﻦ -1أﺣﺴﺐ log10000 log ( x − 1) + log ( x + 3) = 2 -2ﺣﻞ ﻓﻲ -3ﺣﻞ ﻓﻲ
2
x + y = 65 log x + log y = 3
(M