MA129 Lecture 8: Logs Change of Base

• School: WLU
• Department: Mathematics
• Course: MA129
• Professor: Lawrence Howe

Document Summary

2 og (x og x a og 15. + 1 = n 15 / ln 3 log 15 / log 3 log x / log x. Lecture 8 - logs change of base l l. 2log (x+1) x + 1 = 4 x = l ylog 8y = 84/3 = y = 3/4 y = y = 4 y = (8 ) 8 ln (4-x) + ln 2 = 2 ln x. Ln [(4-x)2] = ln x2 eln[(4 x)2] = lnx2 ( x. 8 x2 + x ( 2 x = 4 is not a solution because it does not satisfy the original equation. Clicker l l x ( + 1 (x. 4(0. 12) x x = x ln [4(0. 12) ] ln0. 10 x = ln4 ln0. 12 x ln4 xln0. 12 xln0. 12 xln0. 10 x(ln0. 12 ln0. 10) ln4) / (ln0. 12 (ln5/4) / (ln0. 12/0. 10) ln0. 10) ln2. 2. 58 ln6 ln6/ ln2 n [ 5 (0. 10) ] x x = x =