(log4^3+log8^3)*(log3^2+log9^2)-log1/2^(4√32)（那个4是根号上的）（log旁

首先解释一下ln4=2ln2
底数也有类似规律
log4^3=1/2（log2^3）
所以
(log3^2 + log9^4)×(log4^3 + log8^3)
= (log3^2 + log3^2)×(（1/2）log2^3 + （1/3）log2^3)
=5/3

(log4^3+log8^3)*(log3^2+log9^2)-log2^(4√32)
=(log4^3*8^3)*(log3^2*9^2)-log2^(4√32)
=(log2^15)*(log3^6)-log2^(4√32)
=90*log2*log3 -(16√2)log2

(log3^2 + log9^2)×（log4^3 + log8^3)
＝(log3^2+0.5log3^2)×(0.5log2^3+1/3*log2^3)
=3/2*log3^2×5/6*log2^3
=5/4

题目请描述规范、准确.如,log3^2应该是log9,而不是（log3）^2吧?那么log3^3^1/2是不是log[27^(1/2)]?还有ln根号下e-lg1,指的是ln[(e-lg1)^(1/2)]呢,还是ln[e^(1/2)]-lg1?不说清楚结果是不一样的.

化简：（log4^3+log8^3)(log3^2+log9^2)-log1/2^4根号32
（log4^3+log8^3)(log3^2+log9^2)-log1/2^4根号32
=（log2^6+log2^6)(log3^2+log3^4)-log2^（-4）4√2
=（2*6log2)(6log3)-log2^（-4）2^(5/2)
=12(log2)(6log3)-log2^（-3/2）
=12(log2)(6log3)+3/2log2
y=根号下log1/2^(2X-1) 的定义域
y=√log1/2^(2X-1)
∵log1/2^(2X-1) ≥0
log1/2^(2X-1) ≥log1
1/2^(2X-1) ≥1
2^(2X-1) ≤1=2^0
2X-1≤0
∴x≤1/2
吉林 汪清LLX

10^2a=4x4=16=16lg1010^2a-B=16lg10-lg5=16lg(2x5)-lg5=16lg2+15lg5log4^3+log8^3=log2^6+log2^9=(6+9)log2=15log2log3^2+log9^2=log3^2+log3^4=(2+4)log3=6log3（log4^3+log8^3)x（log3^2+log9^2)=15log2x6log3=90l...

1,换底公式：loga^b=logc^b/logc^a.
loga^b=1/logb^a.
所以有：
log3^2+log9^2
=log3^2+log3^2/log3^9
=log3^2+1/2log3^2=3/2log3^2
=(3/2)/log2^3.
log4^3+log8^3
=log2^3/log2^4+log2^3/log2^8
=1/2log2^3+1/3log2^3
=5/6log2^3.
所以原式=5/4log2^3/log2^3=5/4.
2,公式：lg1/a=-lga.
1/5(lg32+log4^16+6lg1/2)+1/5lg1/5
=1/5(2+lg32-6lg2)-1/5lg5
=1/5(2+lg32-lg64-lg5)
=1/5(2+lg1/2-lg5)
=1/5(2-lg2-lg5)
=1/5(2-lg10)
=1/5.
3,a-1/a=1.
则：(a²+a-²-3)= (a²+a-²-2-1)=(a²-2a•1/a +a-²-1)=(a-1/a) ²-1=1-1=0
所以(a³+a-³)(a²+a-²-3)/(a^4-a^-4) =0.
4,f(x)在[0,+∞)上单调递增,且f(1) ＜f(lgx),所以 1 ＜lgx.
则解不等式得：
x＞10.
即x∈(10,+∞).
f(x)在(-∞,0]上单调递减,则f(-1) ＜f(lgx),所以lgx＜-1.
解不等式得：
0＜x＜1/10.
∴x∈(0,1/10)∪(10,+∞).

因为log1/3^3/4=-log3^3/4=-(og3^3-log3^4)=log3^4-1=log3^(3＋1)-1=log3^3*log3^1-1=-1,所以log4^3>log1/3^3/4.