已知log(32)9=p,log(27)25=q,适用p,q表示lg5p=log32(9)=2/5×log2(3)=2/5×1/log3(2)∴log3(2)=2/5pq=log27(25)=2/3×log3(5)∴log3(5)=3q/2lg5=log3(5)/log3(10)=log3(5)/[log3(2)+log3(5)]=15pq/(4+15pq)结果怎么出来的?
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/05 15:00:35
![已知log(32)9=p,log(27)25=q,适用p,q表示lg5p=log32(9)=2/5×log2(3)=2/5×1/log3(2)∴log3(2)=2/5pq=log27(25)=2/3×log3(5)∴log3(5)=3q/2lg5=log3(5)/log3(10)=log3(5)/[log3(2)+log3(5)]=15pq/(4+15pq)结果怎么出来的?](/uploads/image/z/5407962-42-2.jpg?t=%E5%B7%B2%E7%9F%A5log%EF%BC%8832%EF%BC%899%3Dp%2Clog%EF%BC%8827%EF%BC%8925%3Dq%2C%E9%80%82%E7%94%A8p%2Cq%E8%A1%A8%E7%A4%BAlg5p%3Dlog32%289%29%3D2%2F5%C3%97log2%283%29%3D2%2F5%C3%971%2Flog3%282%29%E2%88%B4log3%282%29%3D2%2F5pq%3Dlog27%2825%29%3D2%2F3%C3%97log3%285%29%E2%88%B4log3%285%29%3D3q%2F2lg5%3Dlog3%285%29%2Flog3%2810%29%3Dlog3%285%29%2F%5Blog3%282%29%2Blog3%285%29%5D%3D15pq%2F%284%2B15pq%29%E7%BB%93%E6%9E%9C%E6%80%8E%E4%B9%88%E5%87%BA%E6%9D%A5%E7%9A%84%3F)
已知log(32)9=p,log(27)25=q,适用p,q表示lg5p=log32(9)=2/5×log2(3)=2/5×1/log3(2)∴log3(2)=2/5pq=log27(25)=2/3×log3(5)∴log3(5)=3q/2lg5=log3(5)/log3(10)=log3(5)/[log3(2)+log3(5)]=15pq/(4+15pq)结果怎么出来的?
已知log(32)9=p,log(27)25=q,适用p,q表示lg5
p=log32(9)=2/5×log2(3)=2/5×1/log3(2)
∴log3(2)=2/5p
q=log27(25)=2/3×log3(5)
∴log3(5)=3q/2
lg5=log3(5)/log3(10)=log3(5)/[log3(2)+log3(5)]=15pq/(4+15pq)
结果怎么出来的?
已知log(32)9=p,log(27)25=q,适用p,q表示lg5p=log32(9)=2/5×log2(3)=2/5×1/log3(2)∴log3(2)=2/5pq=log27(25)=2/3×log3(5)∴log3(5)=3q/2lg5=log3(5)/log3(10)=log3(5)/[log3(2)+log3(5)]=15pq/(4+15pq)结果怎么出来的?
p=log32(9)=2/5×log2(3)=2/5×1/log3(2) 换底公式的推论
∴log3(2)=2/5p
q=log27(25)=2/3×log3(5) 换底公式的推论
∴log3(5)=3q/2
lg5=log3(5)/log3(10)=log3(5)/[log3(2)+log3(5)]=15pq/(4+15pq) 换底公式
公式:loga(N)=logb(N)/[logb(a)]
推论:loga^m (N^n)=(n/m)*loga(N)
神式。