log√3(底)2/3-log√3(底)6

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log√3(底)2/3-log√3(底)6
log√3(底)2/3-log√3(底)6

log√3(底)2/3-log√3(底)6
log√3(底)2/3-log√3(底)6
=log√3(底)(2/3÷6)
=log√3(底)(1/9)
=log√3(底)(1/√3)四次方
=-4

log√3(底)2/3-log√3(底)6=log√3(底)(2/3)/6=log√3(底)(1/9)=log√3(底)(√3)^-4=-4

log√3(底)2/3-log√3(底)6 1、log下标3 2(log下标4 3+log下标8 3)2、log下标1/3 √27+log下标27 1/3 计算:(1) log(2)(3)*log(3)(4)*log(4)(5)(2)log(2)(7)*log(√7)(8) log以4为底3×log以9为底2+log以2为底^4√64 求值log以9为底32×log以64为底27+log以9为底2×log以4为底√27 求值 [log(4)3+log(8)3][log(3)2+log(9)2]-log(1/2)32∧1/4过程详细点,log(4)3是log以4为底3的对数, log(2)5×log(5)81×log(3)4+In√e.还有(log(2)3+log(4)9)(log(3)4+1/2log(3)2) 计算log以2为底25·log以3为底2√2·log以5为底9的结果为 . log(2)9log(3)4等于 log(2)9×log(3)4 4^[log(2)3+log(4)8] 关于数学对数的换底公式推论的问题已知 log(2)(3) = a,log(3(7)=b,用a,b表示log(42)(56)因为log(2)(3)=a,则1/a=log(3)(2),又∵log(3)(7)=b,∴log(42)(56)=log(3)(56)/log(3)(42)=log(3)(7)+3·log(3)(2)/log(3)(7)+log(3)(2)+1=ab+3/ab+b+1 关于数学对数的换底公式推论的问题已知 log(2)(3) = a,log(3(7)=b,用a,b表示log(42)(56)因为log(2)(3)=a,则1/a=log(3)(2),又∵log(3)(7)=b,∴log(42)(56)=log(3)(56)/log(3)(42)=log(3)(7)+3·log(3)(2)/log(3)(7)+log(3)(2)+1=ab+3/ab+b+1 利用对数的换底公式化简下列 1.log(a)c×log(c)a 2.log(2)3×log(3)4×log(4)5×log(5)2 2log底3x+log底3 2=1/2log底3 16 x为什么值 化简1,log(√3)3√3 2,2log(5)10+log(5)0.25 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 log₂3×log₃4+﹙√3﹚∧log₃4=