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發表 studentallen 於 星期日 四月 01, 2007 10:45 am

#ed_op#DIV#ed_cl#Q3:#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#I think the question should be#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#已知 log#ed_op#SUB#ed_cl#k#ed_op#/SUB#ed_cl#(2a-b)=5-log#ed_op#SUB#ed_cl#k#ed_op#/SUB#ed_cl#b 且 log#ed_op#SUB#ed_cl##ed_op#FONT color=#ff0000#ed_cl#c#ed_op#/FONT#ed_cl##ed_op#/SUB#ed_cl#(a#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#+b#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#+8a+6b)=2    .......紅色部份是修改的 後面的條件應該沒問題#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#1. 已知 log#ed_op#SUB#ed_cl#k#ed_op#/SUB#ed_cl#(2a-b)=5-log#ed_op#SUB#ed_cl#k#ed_op#/SUB#ed_cl#b    整理得=>log#ed_op#SUB#ed_cl#k#ed_op#/SUB#ed_cl#b(2a-b)=5    =>b#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#-2ab+k#ed_op#SUP#ed_cl#5#ed_op#/SUP#ed_cl#=0------<1>#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#2.因為b是整數    故式<1>可視為以b為未知的二次方程  #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#   其判別式D=4a#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#-4k#ed_op#SUP#ed_cl#5#ed_op#/SUP#ed_cl#大於等於0    => k#ed_op#SUP#ed_cl#5#ed_op#/SUP#ed_cl#小於等於a#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#BR#ed_cl#3.由5<a<15  => k#ed_op#SUP#ed_cl#5#ed_op#/SUP#ed_cl#<15#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#<3#ed_op#SUP#ed_cl#5    #ed_op#/SUP#ed_cl#因為k是底數  => k不等於0和1   又k是整數 #ed_op#FONT color=#0000ff#ed_cl#k只能是2#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#4.代回<1>得   b(2a-b)=2#ed_op#SUP#ed_cl#5 #ed_op#/SUP#ed_cl#    因為a,b是整數  =>b必為2的正整數次方  #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#                     又5<b<15     => #ed_op#FONT color=#0000ff#ed_cl#b=8#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#5. b=8  8(2a-8)=32  #ed_op#FONT color=#0000ff#ed_cl#a=6#ed_op#/FONT#ed_cl#     =>c#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#=a#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#+b#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#+8a+6b=196   又底數c>0  =>  #ed_op#FONT color=#0000bf#ed_cl#c=14#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#                                                        #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

發表 galaxylee 於 星期日 四月 01, 2007 1:53 am

#ed_op#DIV#ed_cl##ed_op#DIV#ed_cl#2.#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#底下log皆以1/2為底#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#a^(2^a)=1#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#log[a^(2^a)]=1#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#=>(2^a)(loga)=1#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#=>loga=(1/2)^a#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#a是y=logx和y=(1/2)^x圖形交點x坐標#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#因為y=logx和y=(1/2)^x和y=x交於同一點#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以a亦滿足a=loga#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#用反證法#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(1)若a≧1/√2,則1/√2≦a=loga≦log(1/√2)=1/2,矛盾#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(2)若a≦1/2,則a=loga≧log(1/2)=1,矛盾#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#由此得證,1/2<a<1/√2#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#3.#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#b(2a-b)=k^5#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#a^2+b^2+8a+6b=k^2#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#k是不為1的正整數,k^5>k^2#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#=>a^2+b^2+8a+6b<b(2a-b)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#=>a^2+(8-2b)a+(2b^2+6b)<0....(*)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#判別式=(8-2b)^2-4(2b^2+6b)=4[-(b+7)^2+65]#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#當7≦b≦13,判別式<0#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#表示a^2+(8-2b)a+(2b^2+6b)>0恆正#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(*)無a的實數解#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#此題無解,或者題目本身就有問題#ed_op#/DIV#ed_cl##ed_op#/DIV#ed_cl#

發表 studentallen 於 星期六 三月 31, 2007 11:04 pm

#ed_op#DIV#ed_cl##ed_op#P class=MsoNormal style="MARGIN: 0cm 0cm 0pt; mso-pagination: widow-orphan"#ed_cl##ed_op#SPAN style="FONT-FAMILY: 新細明體; mso-font-kerning: 0pt; mso-bidi-font-family: 新細明體"#ed_cl#第二題#ed_op#SPAN lang=EN-US#ed_cl#:#ed_op#?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" /#ed_cl##ed_op#o:p#ed_cl##ed_op#/o:p#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/P#ed_cl##ed_op#P class=MsoNormal style="MARGIN: 0cm 0cm 0pt; mso-pagination: widow-orphan"#ed_cl##ed_op#SPAN lang=EN-US style="FONT-FAMILY: 新細明體; mso-font-kerning: 0pt; mso-bidi-font-family: 新細明體"#ed_cl#1. #ed_op#/SPAN#ed_cl##ed_op#SPAN style="FONT-FAMILY: 新細明體; mso-font-kerning: 0pt; mso-bidi-font-family: 新細明體"#ed_cl#等式#ed_op#SPAN lang=EN-US#ed_cl#  a#ed_op#SUP#ed_cl#2^a #ed_op#/SUP#ed_cl#=0.5  #ed_op#/SPAN#ed_cl#取以#ed_op#SPAN lang=EN-US#ed_cl#0.5#ed_op#/SPAN#ed_cl#為底的對數#ed_op#SPAN lang=EN-US#ed_cl# =>得 2#ed_op#SUP#ed_cl#a#ed_op#/SUP#ed_cl##ed_op#/SPAN#ed_cl##ed_op#SPAN lang=EN-US#ed_cl# log#ed_op#SUB#ed_cl#0.5#ed_op#/SUB#ed_cl#a=1  =>log#ed_op#SUB#ed_cl#0.5#ed_op#/SUB#ed_cl#a=0.5#ed_op#SUP#ed_cl#a#ed_op#o:p#ed_cl##ed_op#/o:p#ed_cl##ed_op#/SUP#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/P#ed_cl##ed_op#P class=MsoNormal style="MARGIN: 0cm 0cm 0pt; mso-pagination: widow-orphan"#ed_cl##ed_op#SPAN lang=EN-US style="FONT-FAMILY: 新細明體; mso-font-kerning: 0pt; mso-bidi-font-family: 新細明體"#ed_cl#                                                                                           => #ed_op#FONT color=#ff0000#ed_cl#a=0.5#ed_op#SUP#ed_cl#0.5^a#ed_op#SPAN style="COLOR: red"#ed_cl# #ed_op#/SPAN#ed_cl##ed_op#/SUP#ed_cl##ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/P#ed_cl##ed_op#P class=MsoNormal style="MARGIN: 0cm 0cm 0pt; mso-pagination: widow-orphan"#ed_cl##ed_op#SPAN lang=EN-US style="FONT-FAMILY: 新細明體; mso-font-kerning: 0pt; mso-bidi-font-family: 新細明體"#ed_cl##ed_op#FONT color=#ff0000#ed_cl##ed_op#SUP#ed_cl##ed_op#SPAN style="COLOR: red"#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/SUP#ed_cl##ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#SPAN lang=EN-US style="FONT-FAMILY: 新細明體; mso-font-kerning: 0pt; mso-bidi-font-family: 新細明體"#ed_cl##ed_op#SUP#ed_cl##ed_op#SPAN style="COLOR: red"#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/SUP#ed_cl##ed_op#/SPAN#ed_cl##ed_op#SPAN lang=EN-US style="FONT-FAMILY: 新細明體; mso-font-kerning: 0pt; mso-bidi-font-family: 新細明體"#ed_cl#2.  a>0  ,0<0.5<1  => #ed_op#/SPAN#ed_cl##ed_op#SPAN#ed_cl# 0<0.5#ed_op#SUP#ed_cl#a#ed_op#/SUP#ed_cl#<1   =>0.5#ed_op#SUP#ed_cl#1#ed_op#/SUP#ed_cl#<0.5#ed_op#SUP#ed_cl#0.5^a#ed_op#/SUP#ed_cl# <0.5#ed_op#SUP#ed_cl#0#ed_op#/SUP#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/P#ed_cl##ed_op#P class=MsoNormal style="MARGIN: 0cm 0cm 0pt; mso-pagination: widow-orphan"#ed_cl##ed_op#SPAN lang=EN-US style="FONT-FAMILY: 新細明體; mso-font-kerning: 0pt; mso-bidi-font-family: 新細明體"#ed_cl##ed_op#o:p#ed_cl#                                                                      | |#ed_op#/o:p#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/P#ed_cl##ed_op#P class=MsoNormal style="MARGIN: 0cm 0cm 0pt; mso-pagination: widow-orphan"#ed_cl##ed_op#SPAN lang=EN-US style="FONT-FAMILY: 新細明體; mso-font-kerning: 0pt; mso-bidi-font-family: 新細明體"#ed_cl##ed_op#o:p#ed_cl#                                                     => 0.5<     a       <1     =>0.5<0.5#ed_op#SUP#ed_cl#a#ed_op#/SUP#ed_cl#<0.5#ed_op#SUP#ed_cl#0.5#ed_op#/SUP#ed_cl##ed_op#/o:p#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/P#ed_cl##ed_op#P class=MsoNormal style="MARGIN: 0cm 0cm 0pt; mso-pagination: widow-orphan"#ed_cl##ed_op#SPAN lang=EN-US style="FONT-FAMILY: 新細明體; mso-font-kerning: 0pt; mso-bidi-font-family: 新細明體"#ed_cl#                                                                                        =>#ed_op#STRONG#ed_cl#0.5#ed_op#SUP#ed_cl#0.5^0.5#ed_op#/SUP#ed_cl#<#ed_op#FONT color=#0000ff#ed_cl#0.5#ed_op#SUP#ed_cl#0.5^a#ed_op#/SUP#ed_cl##ed_op#/FONT#ed_cl##ed_op#/STRONG#ed_cl##ed_op#FONT color=#0000ff#ed_cl#<0.5#ed_op#SUP#ed_cl#0.5#ed_op#/SUP#ed_cl##ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/P#ed_cl##ed_op#P class=MsoNormal style="MARGIN: 0cm 0cm 0pt; mso-pagination: widow-orphan"#ed_cl##ed_op#SPAN lang=EN-US style="FONT-FAMILY: 新細明體; mso-font-kerning: 0pt; mso-bidi-font-family: 新細明體"#ed_cl#3.#ed_op#FONT color=#ff7f00#ed_cl#0.5#ed_op#SUP#ed_cl#0.5#ed_op#/SUP#ed_cl# =1/根號2 >0.5#ed_op#SUP#ed_cl#0.5^a#ed_op#/SUP#ed_cl#=a#ed_op#/FONT#ed_cl# (由紅字和藍字知)#ed_op#/SPAN#ed_cl##ed_op#/P#ed_cl##ed_op#P class=MsoNormal style="MARGIN: 0cm 0cm 0pt; mso-pagination: widow-orphan"#ed_cl##ed_op#SPAN style="FONT-FAMILY: 新細明體; mso-font-kerning: 0pt; mso-bidi-font-family: 新細明體"#ed_cl##ed_op#SPAN lang=EN-US#ed_cl##ed_op#o:p#ed_cl#4.又由0.5#ed_op#SUP#ed_cl#0.5#ed_op#/SUP#ed_cl#<1  =>#ed_op#STRONG#ed_cl#0.5#ed_op#SUP#ed_cl#0.5^0.5#ed_op#/SUP#ed_cl# >0.5#ed_op#/STRONG#ed_cl##ed_op#SUP#ed_cl##ed_op#STRONG#ed_cl#1#ed_op#/STRONG#ed_cl#    #ed_op#/SUP#ed_cl#=>#ed_op#FONT color=#ff7f00#ed_cl#0.5#ed_op#SUP#ed_cl#0.5^a#ed_op#/SUP#ed_cl#= a >0.5#ed_op#SUP#ed_cl#0.5^0.5#ed_op#/SUP#ed_cl# >0.5#ed_op#/FONT#ed_cl# (由粗體字知)#ed_op#/o:p#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/P#ed_cl##ed_op#P class=MsoNormal style="MARGIN: 0cm 0cm 0pt; mso-pagination: widow-orphan"#ed_cl##ed_op#SPAN style="FONT-FAMILY: 新細明體; mso-font-kerning: 0pt; mso-bidi-font-family: 新細明體"#ed_cl##ed_op#SPAN lang=EN-US#ed_cl##ed_op#o:p#ed_cl#5.由橘色知0.5#ed_op#SUP#ed_cl#0.5 #ed_op#/SUP#ed_cl#> a >0.5   .....###ed_op#/o:p#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/P#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

發表 aa2191943 於 星期一 七月 24, 2006 10:57 pm

#ed_op#DIV#ed_cl#第二題:#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#揣摩題意; 我的想法是須證明:#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#-1 < log#ed_op#SUB#ed_cl#2#ed_op#/SUB#ed_cl#a < -0.5#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(但是我不太會做, 只好提供想法)#ed_op#/DIV#ed_cl#

發表 aa2191943 於 星期一 七月 24, 2006 10:33 pm

#ed_op#DIV#ed_cl##ed_op#DIV#ed_cl#第一題不知是何處的考題, 現在這題幾乎是每本講義必備的對數題目; #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(不過我現在是自己想, 沒去翻講義, 也不確定正不正確)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#log x = a + log A#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#log (10/x)= b + log B#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#其中a,b為整數, A,B=[1,10)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#→ 1 = (a+b) + log(AB)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#→ a+b = 0或1#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#討論:#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#1. a+b = 0, a#ed_op#SUP#ed_cl#2 #ed_op#/SUP#ed_cl#- 2b#ed_op#SUP#ed_cl#2 #ed_op#/SUP#ed_cl#= -a#ed_op#SUP#ed_cl#2 #ed_op#/SUP#ed_cl#< 0#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#2. a+b = 1, a#ed_op#SUP#ed_cl#2 #ed_op#/SUP#ed_cl#- 2b#ed_op#SUP#ed_cl#2 #ed_op#/SUP#ed_cl#= (1-b)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# -2b#ed_op#SUP#ed_cl#2 #ed_op#/SUP#ed_cl#= -b#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#-2b+1= -(b+1)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#+2#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#a=2, b=-1, max=2 (此時 x=100)#ed_op#/DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

[數學]3道對數題

發表 宇智波鼬 於 星期一 七月 24, 2006 9:55 pm

1.已知logx及log(10/x)的首數分別為a及b. 試求a^2-2b^2的最大植.

2.已知a>0,且, 求證:


3.已知


a.b.c.k都是整數. 且15>c>b>a>5. 求a.b.c.k的值.