מתמטיקה - 4 יח"ל - כיתה י"ב - חלק א'
ﺟﺑﺭ ﻭﻣﻘﺩّﻣﺔ ﻟﻠﺗّﺣﻠﻳﻝ ﺍﻟﺭّﻳﺎﺿﻲ ﻟﻠﺩّﻭﺍﻝ ﺍﻷﺳّﻳّﺔ ﻭﺍﻟﻠّﻭﻏﺭﻳﺛﻣﻳّﺔ
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ﺍﻟﻔﺻﻝ :5 ﺍﻟﻠّﻭﻏﺭﻳﺛﻣﺎﺕ
) 6 ( ﻟﻮﻏﺮﻳﺜﻢ ﻗﺴﻤﺔ
a log log _ i x y x
0 y x , , 0 a
1
0
log
y
a
a
ﺑﻜﻠﻤﺎت: ﻟﻮﻏﺮﻳﺜﻢ ﻗﺴﻤﺔ ﻳﺴﺎﻭﻱ ﻟﻮﻏﺮﻳﺜﻢ ﺍﻟﺒﺴﻂ ﻧﺎﻗﺺ ﻟﻮﻏﺮﻳﺜﻢ ﺍﻟﻤﻘﺎﻡ ) ﺟﻤﻴﻊ ﺍﻟﻠّﻮﻏﺮﻳﺜﻤﺎﺕ ﺑﻨﻔﺲ ﺍﻷﺳﺎﺱ .( ﺑﺮھﺎن: ﻧﺮﻣﺰ: a log a u x u x ، a v y
a log
y v
x y _ i _ i u v
a a a k u v
ﻧﻌﻮّﺽ ﻭﻧﺤﺼﻞ ﻋﻠﻰ :
u
v
log
log
a log a
a
a
u v
ﻧﺴﺘﻌﻤﻞ ﺍﻟﺨﺎﺻّﻴّﺔ (3) ﻭﻧﺤﺼﻞ ﻋﻠﻰ :
lo
g
o l g log x
y
a
a
a
98 2 _ i
7 7 98 log log log 2
(1)
أﻣﺜﻠﺔ :
log
49 2
7
7
1 ,000
1 2
3
1 1 2 2 2
(2)
log
log
10 log log log 10
1,000
10 3
1 0
96 3 96 3 log log log ( )
ﺃﻱ :
) ﺃ (
log log x
a log (
)
اِﻧﺘﺒﮫﻮا:
y
x y
2
2
2
a
a
log log
log
96
x
96 3
x y
ﺃﻱ :
) ﺏ (
a
2
log
log
log
2
3
a
y
2
a
log log
x
) x y
) ﺝ (
a
a log (
y
a
) 7 ( ﻟﻮﻏﺮﻳﺜﻢ ﻗﻮّة
0 a x ,
b log
b
1
0
log
x
x
a
a
ﺑﻜﻠﻤﺎت: ﻟﻮﻏﺮﻳﺜﻢ ﻗﻮّﺓ ﻳﺴﺎﻭﻱ ﺣﺎﺻﻞ ﺿﺮﺏ ﺃﺱ ﺍﻟﻘﻮّﺓ ﻓﻲ ﻟﻮﻏﺮﻳﺜﻢ ﺃﺳﺎﺱ ﺍﻟﻘﻮّﺓ ) ﺟﻤﻴﻊ ﺍﻟﻠّﻮﻏﺮﻳﺜﻤﺎﺕ ﺑﻨﻔﺲ ﺍﻷﺳﺎﺱ .( ﺑﺮھﺎن: ﻧﺮﻣﺰ:
u
a log
a
x u
x
u _ i
b
b
b
ﻧﻌﻮّﺽ ﻭﻧﺤﺼﻞ ﻋﻠﻰ :
u
log
a log a
a log a
x
a
b b u
ﻧﺴﺘﻌﻤﻞ ﺍﻟﺨﺎﺻّﻴّﺔ (3) ﻭﻧﺤﺼﻞ ﻋﻠﻰ :
v
a log u
log
x
a
a log x . ﺇﺫﺍ ﻛﺎﻥ b ﻫﻮ ﻋﺪﺩ ﺯﻭﺟ ﱞﻲ، ﻳﻜﻮﻥ ﺍﻟﺘّﻌﺒﻴﺮ b x ﻣﻌﺮّﻑ ﻓﻘﻂ ﻟﻜﻞ 0
(1) ﺍﻟﺘّﻌﺒﻴﺮ
اِﻧﺘﺒﮫﻮا:
x ،
ﻣﻌﺮّﻓًﺎ ﻟﻜﻞ 0
a log
x
x ﻳ ﻬﺪﻑ ﺇﻟﻰ ﺍﻟﺘّﺄﻛﻴﺪ ﺑﺄﻧّﻪ ﻻ ﻳﻮﺟﺪ ﺗﻐﻴﻴﺮ ﻓﻲ ﻣﺠﺎﻝ ﺍﻟﺘّﻌﻮﻳﺾ .
ﻭﻟﺬﺍ ﺍﻟﻄّﻠﺐ ﺑﺄﻥ ﻳﻜﻮﻥ 0
x ، ﻭَ b ﻋﺪﺩًﺍ
ﺎ ﺯﻭﺟﻴ .
(2) ﻳﻤﻜﻦ ﺍِﺳﺘﻌﻤﺎﻝ ﺍﻟﻘﺎﻧﻮﻥ (7) ﺃﻳﻀًﺎ ﺇﺫﺍ ﻛﺎﻥ 0
b a a log blog ( ) x x .
ﻟﻜﻦ ﻓﻲ ﻣﺜﻞ ﻫﺬﻩ ﺍﻟﺤﺎﻟﺔ ﻳﺘﺤﻘّﻖ :
3
log
log
log
1 25
5
3
5
3 log 3 2 3 3 2 2 5 log 5
3
أﻣﺜﻠﺔ :
(1)
1.5
log
2 5
3
3
3
1 ,000
1 0
2 .5
log
log
log
log
(2)
10 2.5 10 2.5 1 2.5
0 .5
1 0
1 0
1 3
1 16 4 1
1 3
1 3
3
log
log
(3)
log 16
16
3
2
2
2
4 2 4 4 4 4 2 8 log ( ) log 2
(4)
© ﺟﻣﻳﻊ ﺍﻟﺣﻘﻭﻕ ﻣﺣﻔﻭﻅﺔ ﻟﺟﺎﺑﻲ ﻳﻛﻭﺋﻳﻝ –
– ﺭﻳﺎﺿﻳّﺎﺕ ﻟﻁﻼّﺏ 4 ﻭﺣﺩﺍﺕ ﺗﻌﻠﻳ ﻣﻳّﺔ – ﺍﻟﺻّﻑ ﺍﻟﺛّﺎﻧﻲ ﻋﺷﺭ – ﺍﻟﻣﻧﻬﺞ ﺍﻟﺟﺩﻳﺩ
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