高一数学对数题，来高手帮忙嘞
\u9ad8\u4e00\u6570\u5b66\u5bf9\u6570\u9898\uff0c\u8c01\u6765\u5e2e\u4e0b\u5fd9\u561ea=lg2,b=lg5
lg75=lg25*3=lg25+lg3=2lg5+lg(5-2)=2lg5+lg5/lg2=2b+b/a
\uff081\uff09logaN^n=nlogaN
\u63a8\u5bfc\u516c\u5f0f\uff1a
\u7531\u5bf9\u6570\u52a0\u6cd5\u516c\u5f0f\uff1alogaM+logaN=logaMN\uff0c
n\u4e2alogaN\u76f8\u52a0\u5f97
logaN+ logaN+\u2026+logaN
=loga(N\u00d7N\u00d7\u2026\u00d7N) \uff08n\u4e2aN\u76f8\u4e58\uff09
=logaN^n\uff1b
\uff082\uff09\u5df2\u77e5a=lgx\uff0c\u5219a+3\u7b49\u4e8e\uff1f
a+3=lgx+3=lgx+lg1000=lg1000x\uff1b
\uff083\uff09\u82e52.5^x=1000\uff0c0.25^y=1000\uff0c\u5219(1/x)-(1/y)\u7b49\u4e8e\uff1f
\u22352.5^x=1000\uff0c
\u2234x=log1000\uff0c1/x=log2.5=(lg2.5)/3
(\u8868\u793a\u5e95\u6570\u662f2.5\uff0c\u5176\u4ed6\u7c7b\u4f3c)
\u22350.25^y=1000\uff0c
\u2234y=log1000\uff0c1/y=log0.25=(lg0.25)/3
(1/x)-(1/y)= (lg2.5)/3 - (lg0.25)/3=[(lg2.5) - (lg0.25)]/3=(lg10)/3=1/3\uff1b
\uff084\uff09\u8bbe\u51fd\u6570f\uff08x\uff09=logax\uff08a>0\u4e14a\u4e0d\u7b49\u4e8e0\uff09\uff0c\u82e5f\uff08x1x2\u2026x2010\uff09=8\uff0c
\u5219f\uff08x1²\uff09+f(x2²)+\u2026+f(x2010²)\u7684\u503c\u7b49\u4e8e\uff1f
\u2235f\uff08x1x2\u2026x2010\uff09=8\uff0c
\u2234log( x1x2\u2026x2010)=8\uff0c
\u5373logx1+logx2+\u2026+logx2010=8
f\uff08x1²\uff09+f(x2²)+\u2026+f(x2010²)
= logx1²+logx2²+\u2026+logx2010²
=2 logx1+2logx2+\u2026+2logx2010
=2 (logx1+logx2+\u2026+logx2010)
=2\u00d78
=16.
lg(x+2y)(x-y)=lg2xy
(x+2y)(x-y)=2xy
x^2+xy-2y^2=2xy
x^2-xy-2y^2=0
(x-2y)(x+y)=0
x=2y,x=-y
由定义域，x>0,y>0
所以x=-y不成立
所以x=2y
x/y=2
lg(x-y)+lg(x+2y)=lg(x-y)(x+2y)
lg2+lgx+lgy=lg(2xy)
所以x^-xy-2y^=0
所以x/y=2或-1
因为x>0 y>0
所以x/y=2
lg(x-y)+lg(x+2y)=lg(x-y)(x+2y)=lg(x^2+xy-2y^2),
lg2+lgx+lgy=lg(2xy).所以
x^2+xy-2y^2＝2xy，
因式分解（x+y）(x-2y)=0,
若x+y＝0，则xy<=0,lg(2xy)无意义。
所以x-2y＝0，x/y＝2
绛旓細lgx+lgy=2 lg(xy)=2 xy=100 涓攛>0 y>0 1/x+1/y =(x+y)/(xy)=(x+y)/100 >=(2鈭歺y)/100 =20/100 =1/5 鍙栫瓑鍙锋椂x=y=10
绛旓細log锛坅锛夛紙b锛壝條og锛坆锛夛紙a锛=锛坙gblga锛/锛坙galgb锛=1
绛旓細2.璁緇oga^2=logb^16=x 鍒檃^x=2锛宐^x=16=2^4=(a^x)^4=(a^4)^x 鎵浠=a^4
绛旓細棣栧厛鍛婅瘔浣犲叕寮廇=log锛坅涓哄簳b鐨瀵规暟锛夊彲浠ュ緱鍒癆=log锛坈涓哄簳b鐨勫鏁帮級/log锛坈涓哄簳b鐨勫鏁帮級a=log(3涓哄簳A鐨勫鏁) 鎵浠=lnA/ln3 1/a=ln3/lnA 鍚岀悊寰楋細b=lnA/ln5 1/b=ln5/lnA 鎵浠1/a+1/b=(ln3+ln5)/lnA=2 ln3+ln5=ln15 ln15=2lnA A^2=15 A=鏍瑰彿15 ...
绛旓細1.a=lg(8/7)=lg8-lg7=3lg2-lg7 b=lg(50/49)=5lg2-2lg7 5a=15lg2-5lg7 3b=15lg2-6lg7 lg7=5a-3b 2.3鐨刟娆℃柟=4鐨刡娆℃柟=36 a=lg36/lg3 b=lg36/lg4 2/a+1/b=2lg3/lg36+lg4/lg36=(2lg3+lg4)/lg36=lg36/lg36=1 ...
绛旓細log(a) b (a涓轰笅鏍)杩欎釜瀵规暟涓紝a灏辨槸搴曟暟锛宐鍙仛鐪熸暟锛岀粨鏋滃彨鍋氱湡鏁癰鐨勫鏁 搴曟暟鐨瀵规暟锛鐨勬剰鎬濇槸锛屼互搴曟暟涓虹湡鏁帮紝鍗砽og(a) a=1 杩欎釜濂界悊瑙ｅ惂
绛旓細瑙ｏ細(1)浠=log a X,鍒檟=a^t 鎵浠(t)=[a/(a^2-1)]*[a^t-1/(a^t)]鎵浠(x)=[a/(a^2-1)]*[a^x-1/(a^x)](2)浠诲彇x灞炰簬R,鍒-x灞炰簬R f(-x)=[a/(a^2-1)]*{a^(-x)-1/[a^(-x)]} =[a/(a^2-1)]*[1/(a^x)-a^x]=-f(x)鎵浠(x)鏄鍑芥暟...
绛旓細浠=0,n=1 鍒檉(0/1)=f(0)-f(1)=f(0)寰梖锛1锛=0鍚庨潰鐨勪笉浼氫簡锛屼涪涓嬩袱骞村崐浜嗭紝閮藉繕浜嗭紝鍙兘瑙ｅ喅杩欑偣浜嗭紝sorry 锛2锛変换鍙栦换鎰忕殑m>n>0 鍒檉(m)-f(n)=f(m/n)鍥犱负m>n锛宮/n>1 鎵浠(m/n)>0 鎵浠ュ浠绘剰鐨刴>n>0锛岄兘鏈塮(m)>f(n)鎵浠(x)鏄鍑芥暟銆
绛旓細鍒 log2a=1鎴0 瑙ｅ緱a=2鎴朼=1(涓嶅悎棰樻剰锛岃垗)灏哸=2浠ｅ叆鈶犲紡锛屽彲寰梜=2 f(log2x)>f(1)>log2锛坒x锛夊嵆涓 (log2x)²-(log2x)+2>1²-1+2>log2銆攛²-x+2銆曞嵆涓猴紙log2x)²-(log2x)+2>2---鈶 涓2>log2銆攛²-x+2銆---鈶 瑙ｂ憽寮忓彲寰楋紙...
绛旓細鍦ㄥ悧锛熸湜閲囩撼锛佽阿璋紒
