关于数学对数的换底公式推论的问题已知 log(2)(3) = a,log(3(7)=b,用a,b表示log(42)(56)因为log(2)(3)=a,则1/a=log(3)(2),又∵log(3)(7)=b,∴log(42)(56)=log(3)(56)/log(3)(42)=log(3)(7)+3·log(3)(2)/log(3)(7)+log(3)(2)+1=ab+3/ab+b+1
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![关于数学对数的换底公式推论的问题已知 log(2)(3) = a,log(3(7)=b,用a,b表示log(42)(56)因为log(2)(3)=a,则1/a=log(3)(2),又∵log(3)(7)=b,∴log(42)(56)=log(3)(56)/log(3)(42)=log(3)(7)+3·log(3)(2)/log(3)(7)+log(3)(2)+1=ab+3/ab+b+1](/uploads/image/z/3961167-15-7.jpg?t=%E5%85%B3%E4%BA%8E%E6%95%B0%E5%AD%A6%E5%AF%B9%E6%95%B0%E7%9A%84%E6%8D%A2%E5%BA%95%E5%85%AC%E5%BC%8F%E6%8E%A8%E8%AE%BA%E7%9A%84%E9%97%AE%E9%A2%98%E5%B7%B2%E7%9F%A5+log%282%29%283%29+%3D+a%2Clog%283%287%29%3Db%2C%E7%94%A8a%2Cb%E8%A1%A8%E7%A4%BAlog%2842%29%2856%29%E5%9B%A0%E4%B8%BAlog%282%29%283%29%3Da%2C%E5%88%991%2Fa%3Dlog%283%29%282%29%2C%E5%8F%88%E2%88%B5log%283%29%287%29%3Db%2C%E2%88%B4log%2842%29%2856%29%3Dlog%283%29%2856%29%2Flog%283%29%2842%29%3Dlog%283%29%287%29%2B3%C2%B7log%283%29%282%29%2Flog%283%29%287%29%2Blog%283%29%282%29%2B1%3Dab%2B3%2Fab%2Bb%2B1)
关于数学对数的换底公式推论的问题已知 log(2)(3) = a,log(3(7)=b,用a,b表示log(42)(56)因为log(2)(3)=a,则1/a=log(3)(2),又∵log(3)(7)=b,∴log(42)(56)=log(3)(56)/log(3)(42)=log(3)(7)+3·log(3)(2)/log(3)(7)+log(3)(2)+1=ab+3/ab+b+1
关于数学对数的换底公式推论的问题
已知 log(2)(3) = a,log(3(7)=b,用a,b表示log(42)(56)
因为log(2)(3)=a,则1/a=log(3)(2),又∵log(3)(7)=b,
∴log(42)(56)=log(3)(56)/log(3)(42)=log(3)(7)+3·log(3)(2)/log(3)(7)+log(3)(2)+1=ab+3/ab+b+1
因为我都四年多没接触过咯···完全忘咯原理是怎样的···请高手帮帮忙仔细的讲解一下下··谢谢··特别是log(3)(56)/log(3)(42)=log(3)(7)+3·log(3)(2)/log(3)(7)+log(3)(2)+1=ab+3/an+b+1···这里完全不明白用的是什么方法解的···刚刚申请的账号··分比较少··请见谅···
关于数学对数的换底公式推论的问题已知 log(2)(3) = a,log(3(7)=b,用a,b表示log(42)(56)因为log(2)(3)=a,则1/a=log(3)(2),又∵log(3)(7)=b,∴log(42)(56)=log(3)(56)/log(3)(42)=log(3)(7)+3·log(3)(2)/log(3)(7)+log(3)(2)+1=ab+3/ab+b+1
还记不记得公式,㏒(a)(b)与㏒(b)(a)互为倒数,log(2)(3)=a,则1/a=log(3)(2),这是这一步的原因,
㏒(a)(b)=㏒(c)(b)/㏒(c)(a),所以㏒(42)(56)=㏒(3)(56)㏒(3)(42)
㏒(ab)=㏒a+㏒b ,所以log(3)(56)=log(3)(7)+log(3)(8)
㏒a^b(a的b次幂)=b㏒a,所以㏒(3)(8)=3㏒(3)(2)
分母道理一样,你在看看