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{"id":27524,"date":"2018-04-18T16:24:40","date_gmt":"2018-04-18T09:24:40","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=27524"},"modified":"2018-04-18T16:24:40","modified_gmt":"2018-04-18T09:24:40","slug":"chuong-2-bai-tap-trac-nghiem-giai-tich-12-logarit-phan-2","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/chuong-2-bai-tap-trac-nghiem-giai-tich-12-logarit-phan-2\/","title":{"rendered":"Ch\u01b0\u01a1ng 2 -B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m Gi\u1ea3i t\u00edch 12: L\u00f4garit (Ph\u1ea7n 2)"},"content":{"rendered":"

C\u00e2u 1:<\/b>\u00a0T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c log3<\/sub>100 – log3<\/sub>18 – log3<\/sub>50<\/p>\n

A. -3 \u00a0\u00a0\u00a0B. -2 \u00a0\u00a0\u00a0C. 2 \u00a0\u00a0\u00a0D. 3<\/p>\n

<\/div>\n

C\u00e2u 2:<\/b>\u00a0T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c (log2<\/sub>3)(log9<\/sub>4)<\/p>\n

A. 2\/3 \u00a0\u00a0\u00a0B. 1\u00a0\u00a0\u00a0 C. 3\/2\u00a0 \u00a0D. 4<\/p>\n

C\u00e2u 3:<\/b>\u00a0T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c<\/p>\n

\"\"<\/p>\n

A. -2 \u00a0\u00a0\u00a0 B. 2 \u00a0\u00a0\u00a0C. -3loga<\/sub>5 \u00a0\u00a0\u00a0D. 3loga<\/sub>5<\/p>\n

C\u00e2u 4:<\/b>\u00a010log7<\/sup>\u00a0b\u1eb1ng:<\/p>\n

A. 1\u00a0\u00a0\u00a0B. log7<\/sub>10\u00a0\u00a0\u00a0C. 7\u00a0\u00a0\u00a0D. log7<\/p>\n

C\u00e2u 5:<\/b>\u00a0Cho P = log3<\/sub>(a2<\/sup>b3<\/sup>) (a,b l\u00e0 c\u00e1c s\u1ed1 d\u01b0\u01a1ng). Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u00fang ?<\/p>\n

A. P = 6lpg3<\/sub>a.log3<\/sub>b\u00a0\u00a0\u00a0B. P = 2log3<\/sub>a + 3log3<\/sub>b<\/p>\n

C. P = (1\/2)log3<\/sub>a + (1\/3)log3<\/sub>b \u00a0\u00a0\u00a0D. P = (log3<\/sub>a)2<\/sup>.(log3<\/sub>b)3<\/sup><\/p>\n

C\u00e2u 6:<\/b>\u00a0\u0110\u1eb7t a = log2<\/sub>7, b = log2<\/sub>3. T\u00ednh log2<\/sub>(56\/9) theo a v\u00e0 b<\/p>\n

A. P = 3 + a – 2b \u00a0\u00a0\u00a0B. P = 3 + a – b2<\/sup>\u00a0\u00a0\u00a0C. P = 3a\/2b \u00a0\u00a0\u00a0D. 3a\/b2<\/sup><\/p>\n

C\u00e2u 7:<\/b>\u00a0Bi\u1ebft y = 23x<\/sup>. H\u00e3y bi\u1ec3u th\u1ecb x theo y<\/p>\n

\"\"<\/p>\n

H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n\n\n\n
1-B<\/td>\n2-B<\/td>\n3-A<\/td>\n4-C<\/td>\n5-B<\/td>\n6-A<\/td>\n7-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

C\u00e2u 1:<\/b><\/p>\n

log3<\/sub>100 – log3<\/sub>18 – log3<\/sub>50<\/p>\n

\"\"<\/p>\n

C\u00e2u 2:<\/b><\/p>\n

(log2<\/sub>3)(log9<\/sub>4) = (log2<\/sub>3) = (log32<\/sup><\/sub>22<\/sup>) = (log2<\/sub>3)(log3<\/sub>2) = 1<\/p>\n

C\u00e2u 3:<\/b><\/p>\n

\"\"<\/p>\n

C\u00e2u 4:<\/b><\/p>\n

S\u1eed d\u1ee5ng c\u00f4ng th\u1ee9c aloha<\/sub>b<\/sup><\/p>\n

C\u00e2u 5:<\/b><\/p>\n

P = log3<\/sub>a2<\/sup>\u00a0+ log3<\/sub>b3<\/sup>\u00a0= 2log3<\/sub>a + 3log3<\/sub>b<\/p>\n

C\u00e2u 6:<\/b><\/p>\n

P = log2<\/sub>56 – log2<\/sub>9 = log2<\/sub>(8.7) – log2<\/sub>32<\/sup>\u00a0= log2<\/sub>23<\/sup>\u00a0+ log2<\/sub>7 – 2log2<\/sub>3 = 3 + log2<\/sub>7 – 2log2<\/sub>3 = 3 + a – 2b<\/p>\n

C\u00e2u 7:<\/b><\/p>\n

y = 23x<\/sup>\u00a0<=> 3x = log2<\/sub>y <=> x = (1\/3)log2<\/sub>y<\/p>\n","protected":false},"excerpt":{"rendered":"

C\u00e2u 1:\u00a0T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c log3100 – log318 – log350 A. -3 \u00a0\u00a0\u00a0B. -2 \u00a0\u00a0\u00a0C. 2 \u00a0\u00a0\u00a0D. 3 C\u00e2u 2:\u00a0T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c (log23)(log94) A. 2\/3 \u00a0\u00a0\u00a0B. 1\u00a0\u00a0\u00a0 C. 3\/2\u00a0 \u00a0D. 4 C\u00e2u 3:\u00a0T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A. -2 \u00a0\u00a0\u00a0 B. 2 \u00a0\u00a0\u00a0C. -3loga5 \u00a0\u00a0\u00a0D. 3loga5 C\u00e2u 4:\u00a010log7\u00a0b\u1eb1ng: A. […]<\/p>\n","protected":false},"author":3,"featured_media":27525,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[157],"tags":[1447,1446],"yoast_head":"\nCh\u01b0\u01a1ng 2 -B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m Gi\u1ea3i t\u00edch 12: L\u00f4garit (Ph\u1ea7n 2)<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/lop12.edu.vn\/chuong-2-bai-tap-trac-nghiem-giai-tich-12-logarit-phan-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Ch\u01b0\u01a1ng 2 -B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m Gi\u1ea3i t\u00edch 12: L\u00f4garit (Ph\u1ea7n 2)\" \/>\n<meta property=\"og:description\" content=\"C\u00e2u 1:\u00a0T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c log3100 – log318 – log350 A. -3 \u00a0\u00a0\u00a0B. -2 \u00a0\u00a0\u00a0C. 2 \u00a0\u00a0\u00a0D. 3 C\u00e2u 2:\u00a0T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c (log23)(log94) A. 2\/3 \u00a0\u00a0\u00a0B. 1\u00a0\u00a0\u00a0 C. 3\/2\u00a0 \u00a0D. 4 C\u00e2u 3:\u00a0T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c A. -2 \u00a0\u00a0\u00a0 B. 2 \u00a0\u00a0\u00a0C. -3loga5 \u00a0\u00a0\u00a0D. 3loga5 C\u00e2u 4:\u00a010log7\u00a0b\u1eb1ng: A. 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