Question

高一数学对数计算

1.（lg√27+lg8+lg√1000）/lg120
2.lg^2 5+lg2lg5+lg20
3.设log8 3=p，log3 5=q，用p、q表示lg5
4.25^(1/3log5 27+log5 8)
求详细解答啊详细解答，小弟对数杯具啊T.T

Answer 1

1.（lg√27+lg8+lg√1000）/lg120
=[（3/2)lg3+3lg2+3/2]/(1+2lg2+lg3)
=3/2.

2.lg^2 5+lg2lg5+lg20
=(1-lg2)^2+lg2(1-lg2)+1+lg2
=2.

3.设log8 3=p，log3 5=q，用p、q表示lg5
换底得lg3/[3(1-lg5)]=p,
lg5/lg3=q,
相乘得lg5/[3(1-lg5)]=pq,
∴lg5=3pq-3pqlg5,
(1+3pq)lg5=3pq,
∴lg5=3pq/(1+3pq).

4.25^(1/3log5 27+log5 8)
=5*[2(log<5>3+log<5>8)]
=[5^(log<5>24)]^2
=24^2=576.