忘得差不多了···2log2^5=?4log2^5=?

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忘得差不多了···2log2^5=?4log2^5=?
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忘得差不多了···2log2^5=?4log2^5=?
忘得差不多了···2log2^5=?4log2^5=?

忘得差不多了···2log2^5=?4log2^5=?
(lg5)^2+(lg2)(lg50)
= (lg5)^2+(lg2)(lg25×2)
= (lg5)^2+(lg2)[lg(5^2)+lg2]
= (lg5)^2+(lg2)(2lg5+lg2)
= (lg5)^2+2(lg5)(lg2)+(lg2)^2
= (lg5+lg2)^2
= [lg(5×2)]^2
= (lg10)^2
= 1^2
= 1
log3*(√3/3)+log8*(4)
= [lg(√3/3)]/lg3+lg4/lg8
= (lg√3-lg3)/lg3+(lg2^2)/(lg2^3)
= [lg3^(1/2)-lg3]/lg3+(2lg2)/(3lg2)
= [(1/2)lg3-lg3]/lg3+2/3
= (1/2)-1+2/3
= 2/3-1/2
= 1/6
4^[log2*(5)]+2×5^[log25*(4)]
= (2^2)^[log2*(5)]+2×[25^(1/2)]^[log25*(4)]
= 2^2[log2*(5)]+2×25^(1/2)[log25*(4)]
= {2^[log2*(5)]}^2+2×{25^[log25*(4)]}^(1/2)
= 5^2+2×4^(1/2)
= 25+2×2
= 29
写得够清楚了吧同学,因为我实在是无聊,没事可干了.才吃的空……

忘得差不多了···2log2^5=?4log2^5=? 计算：2^log2^32=_______.计算：log2(4^7·2^5)=_______. 1.7^0.2 log2.1 0.9 0.8^2.2 谁大计算器没带 而且忘了差不多了 没办法啊·· 设函数f(x)=log2 (4x)·log2 (2x)的定义域为[1/4,4] 若t=log2 x 求t的取值范围求y=（x）的最大值与最小值,并求出最值时对应的x的值 最好带上说明 第一问老师讲了一点 写到（log2 4+log2 x）（log2 2+log2 x [log2 1]+[log2 2]+[log2 3]+[log2 4]+[log2 5]+...+[log2 1024]=?[x]表示不超过x的最大整数2为底 答案是8204 log2 (2^x-1)·log2 [2^(x+1)-2] log2 (2^x-1)·log2 [2^(x+1)-2] (log5)^2+log2·log50 求函数y=log2 x/2·log2 4x,x∈[1/4,8]的值域 log2^(4^7*2^5)+log2^6-log2^3 已知-3≤log0.5x≤-3/2,求函数f(x)=log2(x/2·log2 x/4)的最大值和最小值 log2（根号2/2）·log2（根号2/4）麻烦写下过程 求值域,log2 x/2·log2 x/4（x∈[1,8]） log2 5/4+log2 5=多少, log的计算 ：log2（20）-log4（25)= log3(2)×log2（5）×log5（3）=?log2[log2(32)-log2(3÷4）+log2log的计算 ：log2（20）-log4（25)= log3(2)×log2（5）×log5（3）=?log2【log2(32)-log2(3÷4）+log2（6）】=?lgx+lg（x+15）=2 ..X lg8*log2(5)*log5(4)得多少? 已知a>1且a^(lgb)=4次根号下2,求log2(ab)的最小值a^(lgb)=2^1/4,两边取以2为底的对数得:log2(a^(lgb))=log2(2^1/4),即lgblog2(a)=1/4,所以lgb=1/4log2(a),而log2(ab)=log2(a)+log2(b)=log2(a)+lgb/lg2,所以log2(ab)=log2(a)+1/4log2(a)lg2, log2(x – 2 ) + 2 = log2(5)log2(x – 2 ) + 2 = log2(5)