64^1/3-(-2/3)^0+log(2)8+(log(8)9)/(log(2)3)
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64^1/3-(-2/3)^0+log(2)8+(log(8)9)/(log(2)3)
64^1/3-(-2/3)^0+log(2)8+(log(8)9)/(log(2)3)
64^1/3-(-2/3)^0+log(2)8+(log(8)9)/(log(2)3)
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=4-1+3+2/3
=20/3
64^1/3-(-2/3)^0+log(2)8+(log(8)9)/(log(2)3)
2^-1*64^2/3-3^log的底数3(8)+log的底数5(4)*log的底数根号2(5)+(1/cos1)^0
已知log(2)[log(3)(log(4) x)] = log(3) [log(4) (log(2) y)]=0,求x+y的值
2log(5 ) 25+3log(2) 64-8log(2) 1 ( )内代表的是以什么为底
2log(5 ) 25+3log(2) 64-8log(2) 1 ( )内代表的是以什么为底
log底数7[log底数3(log底数2X)]=0,那么x^1/2的值是
2log(5)25+3log(2)64=?
计算:1/[log(2)3]+1/[log(5)3]=?
log(3)1+log(2)根号2+lg1000
比较log(1/2)0.3,log(3)0.2的大小
计算:8^(1-log(83))(27^(2/3))-log(32)
log(根号2-1)(3
求值 log【9(下标)】3+log【 9(下标)】27 log【2】(1/2)- log【1/2】2
(log(2)9+log(4)1/9)乘以(log(3)2+log(9)0.5) 括号里是底数.计算这道题.
log(9)x+log(x^2)3=1的解为?原方程可可化为log(3 ²)x+log(x)²3=1 ,∴1/2*log(3)X+1/2*log(x)3=1,∴[log(3)x]²-2log(x)3+1=0,即[log(3)x-1]²=0请问这里的[log(3)x]²-2log(x)3+1=0是怎么推出来的,刚学基础不好,
log∨6 (64)*log∨4 (1/27)*log∨3( 36)
log∨6 (64)*log∨4 (1/27)*log∨3( 36)
log(1/2)3,log(1/3)3,log(1/4)3比大小