Question

数学计算题求解求解求解

求过程求解求解谢谢

Answer 1

建议利用对数的转换公式，把它们全部都化成相同的底来做：

那么，上面的公式就可以化简为：

=log32/log3 * log9/log4 - log(3/4)/log2 + log6/log2

= (5log2)/log3 * (2log3)/(2log2) - (log3 - log4)/log2 + (log2 + log3)/log2

= 5 - log3/log2 + log4/log2 + log2/log2 + log3/log2

= 5 - log3/log2 + 2 + 1 + log3/lgo2

= 8

Answer 2

减号前用换底公式，换成以10为底的对数。
原式=(lg32/lg3)·(lg9/lg4)－[㏒2
3/4 － ㏒2 6]
=(lg2^5/lg3)·(lg3²/lg2²)－㏒2 [(3/4)÷6]
=(5lg2/lg3)·(2lg3/2lg2)－㏒2 (1/8)
=5－㏒2 2^(-3 )
=5－(-3)·㏒2 2
=5+3
=8