Change of Base Formula | Answer Key

University:
Missouri State University
Course:
MTH 136 | Precalculus 1
Academic year:

2019
Views:

365

Pages:

2
Author:

tristepeon32s

Name The Change of Base Formula Date Period Use a calculator to approximate each to the nearest thousandth. 1) log_3(3.3) (log(3.8j))/(log(3)) = 1.08 * 68 2) log_2(30) log(30/(log(2))) = 11.9 * 69 3) log_4(S) (log(5))/(log(q)) = 1.1lo * 1 4) log_2(2.1) (log(2.1))/(log(2)) = 1.07 * 0.41 5) log 3.55 log^ 3.55 log10 ( notnl(255ary) 6) log_6(13) (log(13))/(log(48)) = 1, 41315 7) log_6(40) (log(40))/(log(6)) = 2 * 0.588 8) log_4(3.5) log 35 log^ 4 =0.9037 9) log_2(2.9) (log(7.9))/(log(2)) = 1.536 10) log_6(22) (log(2))/(log(4)) = 1.7251 11) log_7(8.7) (log(3.7))/(tog * 7) = 7.017 12) log_3(62) (log(12))/(185, 3) = 3.75i * 7 13) log_3(4) log 2 *2^ 2 log_2(2 ^ 3) = 2/3 =0.6lolo^ -1 14) in 94 4,5432 15) log, 8.7 (log(8.7))/(log(2)) = 3.121 16) log_9(71) (log(71))/(log(9)) = 1.94 17) log_13(194) 1 c9 194 102 j 13 =7.0538 18) In 14.1 2.6462 19) log_13(1) log,12.9 log,13 = 0.997 20) log_5(10.818) log(10.818) 18g5 =1.4795 21) log_3(189) l * 0g_{103} * (18 ^ 6)/3 4.7712 22) log_16(194) log_194(log(1.9/100)) = 1.9 23) log_5(183) log^ 183 10 g5 =3.23666 24) log_14(2.6) (log(7.6))/(log|1|) = 0 ,3 21 Critical thinking question: (25) Show that log, 1000=log, 10 algebraically. Log Bare "b" * log 2 1000= log_2(log_2(10.1/8)) = 109 1000 123 10092 1000 log2 1000 Slag 10 log2 1000 (Isro) 3.