求证:log2(π)+log5(π)>2
热心网友
log2(π）+log5(π）＝lgπ/lg2+lgπ/lg5＝(lg5lgπ+lg2lgπ）/lg2*lg5＝lgπ（lg2+lg5)/lg2*lg5＝log2（π）*log5（10）因为派大于2￣（3/2）即根号8 10大于5￣（3/2）即根号75所以log2（π）＞1.5 log5（10）＞1.5所以log2(π)+log5(π)2
热心网友
解：log2(π）+log5(π）＝lgπ/lg2+lgπ/lg5=(lg5lgπ+lg2lgπ）/lg2*lg5=lgπ（lg2+lg5)/lg2*lg5=(lgπ/lg2)*lg5
热心网友
1. 原式=1/log(π)2 + 1/log(π)5 =log(π)10/[log(π)2 * 1/log(π)5],2. 分母用基本不等式得：log(π)2 * 1/log(π)54/log(π)10, 又因为log(π)102.