8^(-1/3)+log3(1/27)+log6(5)*[log5(2)+log5(3)]+10^lg3=?

来源：学生作业帮助网编辑：作业帮时间：2024/05/04 04:06:51

8^(-1/3)+log3(1/27)+log6(5)*[log5(2)+log5(3)]+10^lg3=?
8^(-1/3)+log3(1/27)+log6(5)*[log5(2)+log5(3)]+10^lg3=?

8^(-1/3)+log3(1/27)+log6(5)*[log5(2)+log5(3)]+10^lg3=?
8^(-1/3)+log3(1/27)+log6(5)*[log5(2)+log5(3)]+10^lg3
=1/2 + (-3) + log6(5) * log5(2*3) + 3
=1/2 - 3 + 1 +3
=3/2 =1.5
注意用到了对数的几个规律,
1,基本性质 a^(log(a)b)=b
2,换底公式：log(a)(N)=log(b){N}÷log(b){a}

计算log3[(9^2 * 27^(1/3))/3^(log3 234)] 2log3底2-log3底32/9+log3底8(9/4)^1/2-(-9.6)^0-(27/8)^-2/3+(1.5)^-2 求值 x属于[1/9,27],f(x)=log3(x/27)*log3(3x)的最大值 log2(2)+log5(1)-log1/3(1/27)+[3^log3(5)]^2+3^-log3(2)=? 求值:lg2 +log3^27 +lg5 +3^(1-log3^2) +P3^2 +sin (3π /2) 求值:lg2+log3 27+lg5+3^(1-log3 2)+A(2 3)+sin3π/2 log3 5-log3 15+1/2log9 27 的值解方程log3 (9^x+8) = x+log3 (3^x+1 + 6) log3(x^2-3)=1+log3(x-5/3) log3 (1/8) 等于什么 log3(3x-1)=log3(x-1)+log3(3+x)的解是如题已知log（32）9=p,log（27）25=q,适用p,q表示lg5p=log32(9)=2/5×log2(3)=2/5×1/log3(2)∴log3(2)=2/5pq=log27(25)=2/3×log3(5)∴log3(5)=3q/2lg5=log3(5)/log3(10)=log3(5)/[log3(2)+log3(5)]=15pq/(4+15pq)结果怎么出来的? log3(log3 27)=?3都是底数 3/8*log3/8+1/4*log1/4 怎么算不等式log1/3(x^2-8)>log3(1/2x) 求函数f(x)=[(log3,x/27)(log3,3x)]在区间[1,9]上的最大值与最小值计算:log3 8×log4 27+5^(1-log5 2)+(27/8)^-(2/3)+sin3π/2 8^(-1/3)+log3(1/27)+log6(5)*[log5(2)+log5(3)]+10^lg3=?