log2(3)*log3(4)*log4(5)*log5(6)*log6(7)*log7(8)=?
发布网友
发布时间:2024-09-29 07:07
我来回答
共3个回答
热心网友
时间:2024-11-04 12:56
使用换底公式有
log3(4)=log2(4)/log2(3)
log4(5)=log2(5)/log2(4)
依次类推
得到最后一个式子为
log7(8)=log2(8)/log2(7)
各式相乘得
log2(3)*log3(4)*log4(5)*log5(6)*log6(7)*log7(8)
=log2(3)*log2(4)/log2(3)*log2(5)/log2(4)*...*log2(8)/log2(7)
=log2(8)=3
答案还满意吗?
热心网友
时间:2024-11-04 12:56
(log2^9) * (log3^根号下五) * (log5^2)
= [log2^(3^2)] * [log3^(5^1/2)] * [1/(log2^5)]
= 2*log2^3 * 1/2*log3^5 * [1/(log2^5)]
= [(log2^3)/(log2^5)]*(log3^5)
= (log5^3)*(log3^5)
= 1
热心网友
时间:2024-11-04 12:57
log2(3)*log3(4)*log4(5)*log5(6)*log6(7)*log7(8)
=log2(3)*log2(4)/log2(3)*log2(5)/log2(4)*...*log2(8)/log2(7)
=log2(8)=3