高中数学对数计算问题
\u9ad8\u4e2d\u6570\u5b66\u7684\u6240\u6709\u5bf9\u6570\u8ba1\u7b97\u516c\u5f0f \u6025\u554a\u5b9a\u4e49\uff1a
\u3000\u3000\u82e5a^n=b(a>0\u4e14a\u22601)
\u3000\u3000\u5219n=log(a)(b)
\u3000\u3000\u57fa\u672c\u6027\u8d28\uff1a
\u3000\u30001\u3001a^(log(a)(b))=b
\u3000\u30002\u3001log(a)(MN)=log(a)(M)+log(a)(N);
\u3000\u30003\u3001log(a)(M\u00f7N)=log(a)(M)-log(a)(N);
\u3000\u30004\u3001log(a)(M^n)=nlog(a)(M)
\u3000\u3000\u63a8\u5bfc
\u3000\u30001\u3001\u56e0\u4e3an=log(a)(b)\uff0c\u4ee3\u5165\u5219a^n=b\uff0c\u5373a^(log(a)(b))=b\u3002
\u3000\u30002\u3001MN=M\u00d7N
\u3000\u3000\u7531\u57fa\u672c\u6027\u8d281(\u6362\u6389M\u548cN)
\u3000\u3000a^[log(a)(MN)] = a^[log(a)(M)]\u00d7a^[log(a)(N)]
\u3000\u3000\u7531\u6307\u6570\u7684\u6027\u8d28
\u3000\u3000a^[log(a)(MN)] = a^{[log(a)(M)] + [log(a)(N)]}
\u3000\u3000\u53c8\u56e0\u4e3a\u6307\u6570\u51fd\u6570\u662f\u5355\u8c03\u51fd\u6570\uff0c\u6240\u4ee5
\u3000\u3000log(a)(MN) = log(a)(M) + log(a)(N)
\u3000\u30003\u3001\u4e0e\uff082\uff09\u7c7b\u4f3c\u5904\u7406
\u3000\u3000MN=M\u00f7N
\u3000\u3000\u7531\u57fa\u672c\u6027\u8d281(\u6362\u6389M\u548cN)
\u3000\u3000a^[log(a)(M\u00f7N)] = a^[log(a)(M)]\u00f7a^[log(a)(N)]
\u3000\u3000\u7531\u6307\u6570\u7684\u6027\u8d28
\u3000\u3000a^[log(a)(M\u00f7N)] = a^{[log(a)(M)] - [log(a)(N)]}
\u3000\u3000\u53c8\u56e0\u4e3a\u6307\u6570\u51fd\u6570\u662f\u5355\u8c03\u51fd\u6570\uff0c\u6240\u4ee5
\u3000\u3000log(a)(M\u00f7N) = log(a)(M) - log(a)(N)
\u3000\u30004\u3001\u4e0e\uff082\uff09\u7c7b\u4f3c\u5904\u7406
\u3000\u3000M^n=M^n
\u3000\u3000\u7531\u57fa\u672c\u6027\u8d281(\u6362\u6389M)
\u3000\u3000a^[log(a)(M^n)] = {a^[log(a)(M)]}^n
\u3000\u3000\u7531\u6307\u6570\u7684\u6027\u8d28
\u3000\u3000a^[log(a)(M^n)] = a^{[log(a)(M)]*n}
\u3000\u3000\u53c8\u56e0\u4e3a\u6307\u6570\u51fd\u6570\u662f\u5355\u8c03\u51fd\u6570\uff0c\u6240\u4ee5
\u3000\u3000log(a)(M^n)=nlog(a)(M)
\u3000\u3000\u57fa\u672c\u6027\u8d284\u63a8\u5e7f
\u3000\u3000log(a^n)(b^m)=m/n*[log(a)(b)]
\u3000\u3000\u63a8\u5bfc\u5982\u4e0b\uff1a
\u3000\u3000\u7531\u6362\u5e95\u516c\u5f0f\uff08\u6362\u5e95\u516c\u5f0f\u89c1\u4e0b\u9762\uff09[lnx\u662flog(e)(x)\uff0ce\u79f0\u4f5c\u81ea\u7136\u5bf9\u6570\u7684\u5e95]
\u3000\u3000log(a^n)(b^m)=ln(b^m)\u00f7ln(a^n)
\u3000\u3000\u7531\u57fa\u672c\u6027\u8d284\u53ef\u5f97
\u3000\u3000log(a^n)(b^m) = [m\u00d7ln(b)]\u00f7[n\u00d7ln(a)] = (m\u00f7n)\u00d7{[ln(b)]\u00f7[ln(a)]}
\u3000\u3000\u518d\u7531\u6362\u5e95\u516c\u5f0f
\u3000\u3000log(a^n)(b^m)=m\u00f7n\u00d7[log(a)(b)] --------------------------------------------\uff08\u6027\u8d28\u53ca\u63a8\u5bfc \u5b8c\uff09
\u7f16\u8f91\u672c\u6bb5\u51fd\u6570\u56fe\u8c61
\u3000\u30001.\u5bf9\u6570\u51fd\u6570\u7684\u56fe\u8c61\u90fd\u8fc7(1,0)\u70b9.
\u3000\u30002.\u5bf9\u4e8ey=log(a)(n)\u51fd\u6570,
\u3000\u3000\u2460,\u5f530<a<1\u65f6,\u56fe\u8c61\u4e0a\u51fd\u6570\u663e\u793a\u4e3a(0,+\u221e)\u5355\u51cf.\u968f\u7740a \u7684\u589e\u5927,\u56fe\u8c61\u9010\u6e10\u4ee5(1,0)\u70b9\u4e3a\u8f74\u987a\u65f6\u9488\u8f6c\u52a8,\u4f46\u4e0d\u8d85\u8fc7X=1.
\u3000\u3000\u2461\u5f53a>1\u65f6,\u56fe\u8c61\u4e0a\u663e\u793a\u51fd\u6570\u4e3a(0,+\u221e)\u5355\u589e,\u968f\u7740a\u7684\u589e\u5927,\u56fe\u8c61\u9010\u6e10\u4ee5(1.0)\u70b9\u4e3a\u8f74\u9006\u65f6\u9488\u8f6c\u52a8,\u4f46\u4e0d\u8d85\u8fc7X=1.
\u3000\u30003.\u4e0e\u5176\u4ed6\u51fd\u6570\u4e0e\u53cd\u51fd\u6570\u4e4b\u95f4\u56fe\u8c61\u5173\u7cfb\u76f8\u540c,\u5bf9\u6570\u51fd\u6570\u548c\u6307\u6570\u51fd\u6570\u7684\u56fe\u8c61\u5173\u4e8e\u76f4\u7ebfy=x\u5bf9\u79f0.
\u7f16\u8f91\u672c\u6bb5\u5176\u4ed6\u6027\u8d28
\u3000\u3000\u6027\u8d28\u4e00\uff1a\u6362\u5e95\u516c\u5f0f
\u3000\u3000log(a)(N)=log(b)(N)\u00f7log(b)(a)
\u3000\u3000\u63a8\u5bfc\u5982\u4e0b\uff1a
\u3000\u3000N = a^[log(a)(N)]
\u3000\u3000a = b^[log(b)(a)]
\u3000\u3000\u7efc\u5408\u4e24\u5f0f\u53ef\u5f97
\u3000\u3000N = {b^[log(b)(a)]}^[log(a)(N)] = b^{[log(a)(N)]*[log(b)(a)]}
\u3000\u3000\u53c8\u56e0\u4e3aN=b^[log(b)(N)]
\u3000\u3000\u6240\u4ee5 b^[log(b)(N)] = b^{[log(a)(N)]*[log(b)(a)]}
\u3000\u3000\u6240\u4ee5 log(b)(N) = [log(a)(N)]*[log(b)(a)] {\u8fd9\u6b65\u4e0d\u660e\u767d\u6216\u6709\u7591\u95ee\u770b\u4e0a\u9762\u7684}
\u3000\u3000\u6240\u4ee5log(a)(N)=log(b)(N) / log(b)(a)
\u3000\u3000\u516c\u5f0f\u4e8c\uff1alog(a)(b)=1/log(b)(a)
\u3000\u3000\u8bc1\u660e\u5982\u4e0b\uff1a
\u3000\u3000\u7531\u6362\u5e95\u516c\u5f0f log(a)(b)=log(b)(b)/log(b)(a) ----\u53d6\u4ee5b\u4e3a\u5e95\u7684\u5bf9\u6570
\u3000\u3000log(b)(b)=1 =1/log(b)(a) \u8fd8\u53ef\u53d8\u5f62\u5f97: log(a)(b)\u00d7log(b)(a)=1
\u3000\u3000\u5728\u5b9e\u7528\u4e0a\uff0c\u5e38\u91c7\u7528\u4ee510\u4e3a\u5e95\u7684\u5bf9\u6570\uff0c\u5e76\u5c06\u5bf9\u6570\u8bb0\u53f7\u7b80\u5199\u4e3algb,\u79f0\u4e3a\u5e38\u7528\u5bf9\u6570\uff0c\u5b83\u9002\u7528\u4e8e\u6c42\u5341\u8fdb\u4f2f\u5236\u6574\u6570\u6216\u5c0f\u6570\u7684\u5bf9\u6570\u3002\u4f8b\u5982lg10=1,lg100=lg102=2,lg4000=lg\uff08103\u00d74\uff09=3+lg4,\u53ef\u89c1\u53ea\u8981\u5bf9\u67d0\u4e00\u8303\u56f4\u7684\u6570\u7f16\u5236\u51fa\u5bf9\u6570\u8868\uff0c\u4fbf\u53ef\u5229\u7528\u6765\u8ba1\u7b97\u5176\u4ed6\u5341\u8fdb\u5236\u6570\u7684\u5bf9\u6570\u7684\u8fd1\u4f3c\u503c\u3002\u5728\u6570\u5b66\u7406\u8bba\u4e0a\u4e00\u822c\u90fd\u7528\u4ee5\u65e0\u7406\u6570e=2.7182818\u2026\u2026\u4e3a\u5e95\u7684\u5bf9\u6570\uff0c\u5e76\u5c06\u8bb0\u53f7 loge\u3002\u7b80\u5199\u4e3aln\uff0c\u79f0\u4e3a\u81ea\u7136\u5bf9\u6570\uff0c\u56e0\u4e3a\u81ea\u7136\u5bf9\u6570\u51fd\u6570\u7684\u5bfc\u6570\u8868\u8fbe\u5f0f\u7279\u522b\u7b80\u6d01\uff0c\u6240\u4ee5\u663e\u51fa\u4e86\u5b83\u6bd4\u5176\u4ed6\u5bf9\u6570\u5728\u7406\u8bba\u4e0a\u7684\u4f18\u8d8a\u6027\u3002\u5386\u53f2\u4e0a\uff0c\u6570\u5b66\u5de5\u4f5c\u8005\u4eec\u7f16\u5236\u4e86\u591a\u79cd\u4e0d\u540c\u7cbe\u786e\u5ea6\u7684\u5e38\u7528\u5bf9\u6570\u8868\u548c\u81ea\u7136\u5bf9\u6570\u8868\u3002\u4f46\u968f\u7740\u7535\u5b50\u6280\u672f\u7684\u53d1\u5c55\uff0c\u8fd9\u4e9b\u6570\u8868\u5df2\u9010\u6e10\u88ab\u73b0\u4ee3\u7684\u7535\u5b50\u8ba1\u7b97\u5de5\u5177\u6240\u53d6\u4ee3\u3002
log以2为底3的对数*log以3为底4的对数=log以2为底4的对数=2
lg0.01=lg10^(-2)=-2
ln根号e=1/2
2的1+log以2为底3的对数的次方=2*2的log以2为底3的对数的次方=2*3=6
所以答案是 13/2
绛旓細锛1锛2锛塮(x)=lg(3x-2)+2鎭掕繃瀹氱偣,鍒欏彧闇瑕佹妸lg(3x-2)=0鍗冲彲銆傞偅涔3x-2=1,鍒檟=1 姝ゆ椂f(1)=2 鎵浠(x)=lg(3x-2)+2鎭掕繃瀹氱偣锛1锛2锛
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绛旓細瀵规暟鐨杩愮畻娉曞垯锛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 鎸囨暟鐨勮繍绠楁硶鍒欙細1銆乕a^m]脳[a^n]=a^(m锛媙) 銆愬悓搴曟暟骞傜浉涔,搴曟暟涓...
绛旓細1.B 2.1 3.鈭2/2 4.1/3 5.(1)x=2 鏁呭師寮=7/8 (2)浠/n=k锛屽洜涓簃>n,鎵浠>1 鍒檏^2-6k+1=0 k=3+2鈭2 鏈変笉鏄庣櫧鐨勫彲浠ヨ拷闂 鏈涢噰绾
绛旓細棣栧厛鏃犺濡備綍蹇呮湁x^2+x+1/2锛0锛2x^2-x+5/8锛0锛岃仈绔嬭В寰楋細x鈭圧銆傚洜涓篺(x)鏄紑鍙ｅ悜涓嬬殑浜屾鍑芥暟锛屼笖瀵圭О杞翠负x=1锛屾墍浠ュ垎浠ヤ笅涓ょ鎯呭喌杩涜璁ㄨ锛氾紙1锛夊綋log(1/2)(x^2+x+1/2)锛瀕og(1/2)(2x^2-x+5/8)鏃讹紝鍗硏锛1-鈭17/4鎴杧锛1+鈭17/4鏃讹紝鏍规嵁f(x)鍏充簬x=1瀵圭О锛屽繀鐒...
绛旓細锛4锛夎嫢寮忎腑骞傛寚鏁板垯鏈変互涓嬬殑姝ｆ暟鐨勭畻鏈牴鐨瀵规暟杩愮畻娉曞垯:涓涓鏁扮殑绠楁湳鏍圭殑瀵规暟锛岀瓑浜庤寮鏂规暟鐨勫鏁伴櫎浠ユ牴鎸囨暟锛屽嵆锛氳嚜鐒跺鏁颁互甯告暟e涓哄簳鏁扮殑瀵规暟锛岃浣渓nN(N>0)銆鏁板涓篃甯歌浠ogx琛ㄧず鑷劧瀵规暟锛屾墍浠nx鐨勮绠楁柟寮忎篃鍙互鍒╃敤濡備笂鍏紡銆2銆乴n2-ln1鍒╃敤濡備笂鍏紡锛2锛夊緱锛歭n2-ln1=ln锛2/1锛...
绛旓細瀵规暟鐨勬蹇佃嫳璇悕璇嶏細</B>logarithms 濡傛灉a^n=b锛岄偅涔坙og(a)(b)=n銆傚叾涓紝a鍙仛鈥滃簳鏁扳濓紝b鍙仛鈥滅湡鏁扳濓紝n鍙仛鈥滀互a涓哄簳b鐨勫鏁扳濄 log(a)(b)鍑芥暟鍙仛瀵规暟鍑芥暟銆傚鏁板嚱鏁颁腑b鐨勫畾涔夊煙鏄痓>0锛岄浂鍜岃礋鏁版病鏈夊鏁帮紱a鐨勫畾涔夊煙鏄痑>0涓攁鈮1銆 ]瀵规暟鐨勬ц川鍙婃帹瀵煎畾涔夛細鑻^n=b(a>0涓攁鈮...
绛旓細浠涓哄簳y鐨瀵规暟 绛夊悓浜巐ny/lnx 鎵浠ュ師寮忓氨=ln5/ln6 * ln6/ln5 =1
