解
logâ2-1 (3+2â2)
=logâ2-1 (1+â2)^2
=2logâ2-1 (â2+1)
=2logâ2-1 (â2-1)^(-1)
=2x(-1)logâ2-1 (â2-1)
=2x(-1)x1
=-2
æ示ï¼
1ãæ¬é¢åç®è¦è¿ç¨ä»¥ä¸å
¬å¼ï¼loga (b^m)=mloga (b)
2ãç¥é以ä¸å
³ç³»ï¼
ï¼â2+1ï¼ï¼â2-1ï¼=1
å³ â2+1=1/ï¼â2-1ï¼=ï¼â2-1ï¼^(-1)
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