Question

求帮我解答几道高中数学log的题.求详细解答过程。

设log(18)2=a 则log(2)81=? (括号里面是底数)
若log(21)3=a 则log(21)1/49 (就是四十九分之一)
若a=2分之根号2. n=lg2分之lg3，则a的N次方等于?
7的 log(3)7分之3 次方等于?
计算log(2)6乘以log(3)6减去括号log(2)3加上log(3)2收括号.等于
计算括号log(4)3加上log(8)3收括号.乘以括号log(3)2加上log(9)2收括号.等于

抱歉题有些多 等级没到没办法传图片.
能帮我解答几道就解答几道就好.
万分感谢!!
我的QQ2234673058.
看不懂我也可以把图片发给你的^^!

Answer 1

(1)log(18)2=a
18^a=2两边取以2为底的对数
log(2)18^a=1
alog(2)18=1
log(2)2*9=1/a
1+log(2)9=1/a
log(2)9=1/a-1
2log(2)9=2/a-2
log(2)81=2/a-2
(2)log(21)3=a
log(21)1/49=-log(21)49=-log(21)(21/3)^2=-2log(21)21/3
=-2[log(21)21-log(21)3]
=-2+2a=2a-2
(3)a^n=(√2/2)^(lg3/lg2)
=(1/√2)^log(2)3
=2^(-1/2)log(2)3
=2^log(2)3^(-1/2)
=3^(-1/2)
=√3/3
(4)7^log(3)3/7
=7^[log(3)3-log(3)7]
=7^[1-log(3)7]
=7/7^log(3)7
=7/[1/7^log(7)3]
=7/[1/3]
=21
(5)log(2)6*log(3)6-[log(2)3+log(3)2]
=1/log(6)2*1/log(6)3-[log(6)3/log(6)2+log(6)2/log(6)3]
=1/[log(6)2*log(6)3]-[log(6)3*log(6)3+log(6)2*log(6)2]/[log(6)2*log(6)3]
=[1-log(6)3*log(6)3-log(6)2*log(6)2]/[log(6)2*log(6)3]
=[(1+log(6)3)(1-log(6)3)-log(6)2*log(6)2]/[log(6)2*log(6)3]
=[log(6)18*log(6)2-log(6)2*log(6)2]/[log(6)2*log(6)3]
=log(6)2*log(6)9/[log(6)2*log(6)3]
=2[log(6)2*log(6)3]/[log(6)2*log(6)3]
=2
(6)[log(4)3+log8(3)]*[log(3)2+log(9)2]
因log(4)3=1/log(3)4=1/[2log(3)2]=1/2*1/log(3)2=1/2*log(2)3
同理，其他式子也可以按此方法简化
因此原式等于：
[1/2*log(2)3+1/3*log(2)3]*[log(3)2+1/2*log(3)2]
=[5/6*log(2)3]*[3/2*log(3)2]
=5/6*3/2
=5/4

Answer 2

1、log(2)81=log(2)（3^4）=4log(2)3
log(18)2=1/log(2)18=1/(log(2)(9*2))=1/(1+2log(2)3)=a
log(2)3=1/2a-1/2
所以：log(2)81=2/a-2
2、log(21)(1/49)= log(21)(7^-2)=-2 log(21)7=-2( log(21)(21/3))=-2(1-log(21)3)=2a-2
3、a^n=(√2/2)^(lg3/lg2)=(2^(-1/2))^(log(2)3)=2^(-1/2log(2)3)=2^(log(2)(3^(-1/2))=3^(-1/2)=√3/3
4、7^(3/log(3)7)=7^(3log(7)3)=7^(log(7)(3^3))=3^3=27

Answer 3

太多了，这个符号不好打 就写一题不好意思，实际上就是用换地公式与运算法则相结合！

Answer 4

7^(log(3)*(3/7)) = 1.4886962749287
计算括号log(4)3加上log(8)3收括号.乘以括号log(3)2加上log(9)2收括号.等于
=[1/2log(2)3+1/3log(2)3]*[log(3)2+1/2log(3)2]
=5/6log(2)3*3/2log(3)2
=5/4