solve:log(x+10)+log(x-2)=2log(x+3)?
回答 (2)
2015-04-09 3:11 pm
(x+10)(x-2) = (x+3)^2;
x^2 + 8x - 20 = x^2 + 6x + 9;
2x = 29 => x = 14.5
x^2 + 8x - 20 = x^2 + 6x + 9;
2x = 29 => x = 14.5
2015-04-09 3:42 pm
log(x+10) + log(x-2) = 2 log(x+3)
log ( (x+10)(x-2) ) = log (x+3)^2
log ( x^2+10x-2x-20) = log( x^2 + 6x+9)
log ( x^2 +8x - 20) = log (x^2 + 6x + 9)
x^2 + 8x - 20 = x^2 + 6x + 9
8x - 20 = 6x + 9
2x = 29
x = 14.5
log ( (x+10)(x-2) ) = log (x+3)^2
log ( x^2+10x-2x-20) = log( x^2 + 6x+9)
log ( x^2 +8x - 20) = log (x^2 + 6x + 9)
x^2 + 8x - 20 = x^2 + 6x + 9
8x - 20 = 6x + 9
2x = 29
x = 14.5
收錄日期: 2021-04-23 22:21:49
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150409070927AAm2TgM