高一数学对数题
\u9ad8\u4e00\u6570\u5b66\u5bf9\u6570\u8fd0\u7b97\u9898\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.
f(-x)=-f(x)
(1-mx)/(x-1)=((1+mx)/-x-1)^-1
1-m^2 * x^2=1-x^2
m=+ - 1,m=1\u65f6f(x)\u65e0\u610f\u4e49\u820d\u53bb\uff0c\u6545m\uff1d\uff0d1
2.log(√2)(√6+4√2-√6-4√2)
=1/2log(√2)(√6+4√2-√6-4√2)^2
=1/2log(√2)（8）
=3/2log(√2)（2)
=3
第一题：二分之一(a+2b)
绛旓細1銆佽В锛氾紙1锛夊綋a锛1鏃讹紝鍑芥暟f(x)涓哄鍑芥暟锛屾鏃跺嚱鏁癴(2)=2锛屽嵆loga(2²-2+1)=2锛屼篃鍗筹細a²=3锛宎=鈭3 锛2锛夊綋0锛渁锛1鏃讹紝鍑芥暟f(x)涓哄噺鍑芥暟锛屾鏃秄(0)=2锛屽嵆loga(0-0+1)=2锛屽嵆a²=1锛宎=1锛屼笉鍚棰鎰忥紝鏁呰垗鍘 缁间笂鎵杩帮細婊¤冻棰樻剰鐨勫疄鏁癮鐨勫间负2 2銆佽В...
绛旓細鐢ㄦ崲搴曞叕寮弆og3 锛7锛壜穕og2锛 9锛壜穕og49锛2锛= (lg7)/(lg3) 涔樹互(lg9)/(lg2)涔樹互锛坙g2)/(lg49) = (lg7)/(lg3) 涔樹互(lg9)/(lg49)鍥犱负7*7=49,3*3=9鎵浠g49=2lg7,lg9=2lg3 (lg7)/(lg3) 涔樹互(lg9)/(lg49)=1,1,
绛旓細鍘熷紡=lg5/lg2-2(lg10/lg4)=lg5/lg2-lg10/lg2=log2 (//2)=-1 鍘熷紡=lg4+lg5(lg20+lg5)=lg4+2lg5=lg4X25=lg100=2 鍘熷紡=lg2.4=lg24-1=lg3+3lg2-1=0.4771+0.9030-1=0.3801 lg鈭6=1/2lg6=1/2(lg3+lg2)=0.38905 ...
绛旓細閫氳繃鎹㈠簳鍏紡鍙互鎺ㄥ 鎹㈠簳鍏紡 log3c脳logc3锛1 log3c锛漚 甯﹀叆寰 1/a锛漧ogc3 寰楄瘉銆
绛旓細1銆乴og12 5=lg5/lg12=(1-lg2)/(lg3+2lg2)=(1-a)/(b+2a)2銆乴og2 3=a,鈫抣g3/lg2=a---(1),log3 7=b鈫抣g7/lg3=b---(2)(1)*(2):lg7/lg2=ab log14 56=lg56/lg14=(lg7+3lg2)/(lg2+lg7)=(3+lg7/lg2)/(1+lg7/lg2)=(3+ab)/(1+ab)3銆佽В娉1锛15+8+1...
绛旓細浠(x)=log2(x+4)-(1/3)^x 鍒欏洜涓簂og2(x+4), -(1/3)^x閮芥槸澧炲嚱鏁帮紝鎵浠(x)涓哄鍑芥暟锛屽嵆鏈澶氬彧鏈変竴涓浂鐐 鍙坒(-3)=-27<0 f(0)=2-1=1>0 鍥犳鍑芥暟鏈夊敮涓闆剁偣锛屼笖鍦ㄥ尯闂(-3, 0)鍥犳鍘熸柟绋嬪彧鏈変竴涓疄鏍广
绛旓細鑻og3 7路log2 9路log49 a=log4 1/2锛屽垯鏈夛細(lg7/lg3)路(lg9/lg2)路(lga/lg49)=lg(1/2)/lg4 鍗筹細(lg7/lg3)路(2lg3/lg2)路 lga/(2lg7)=-lg2/(2lg2)鍖栫畝寰楋細lga/lg2=-1/2 鍗筹細lga=(-1/2)lg2=-lg鏍瑰彿2=lg(鏍瑰彿2)/2 鎵浠ワ細a=(鏍瑰彿2)/2 ...
绛旓細瀵规暟鏄楂樹竴鏁板蹇呬慨涓瀛︾殑銆傚鏁扮殑杩愮畻娉曞垯锛1銆乴og(a) (M路N锛=log(a) M+log(a) N 2銆乴og(a) (M梅N)=log(a) M-log(a) N 3銆乴og(a) M^n=nlog(a) M 4銆乴og(a)b*log(b)a=1 5銆乴og(a) b=log (c) b梅log (c) a 瀵规暟搴旂敤 瀵规暟鍦ㄦ暟瀛﹀唴澶栨湁璁稿搴旂敤銆傝繖浜涗簨浠朵腑鐨...
绛旓細log(3,2)=lg2/lg3 ln2=lg2/lge 鑰宔=2.718...<3 鎵浠g3>lge>lg2>0 鎵浠n2>log(3,2)鎵浠n2>log(3,2)>log(4,2)=1/2=4^(-1/2)>5^(-1/2)
绛旓細鐢ㄥ埌鐨瀵规暟鍏紡锛歭og(a)(MN)=log(a)(M)+log(a)(N)锛宭og(a^n)M=1/nlog(a)(M) ,loga(a)=1 log8 3=log(2^3)3=1/3log2 3=a/3 log2 6=log2 2*3=log2 2+log2 3=1+a log8 3 鈥 log2 6=a/3-1-a=-1-2a/3 杩欎釜棰樻槸姹備笉鍑烘潵鍏蜂綋鍊肩殑锛屽洜涓篴鐨勫间笉鐭ラ亾鏄灏...
