[log(4)3+log(8)3][log(3)2+log(9)2]-log(1/2)32∧1/4过程详细点,
1个回答
log(4)3+log(8)3
用换底公式,再加loga (x^n)=nlog a x
=ln3/ln4+ln3/ln8
=ln3/ln(2^2)+ln3/ln(2^3)
=ln3/(2ln2)+ln3/(3ln2)
=(ln3/ln2)(1/2+1/3)
=(5/6)(ln3/ln2)
同理
log(3)2+log(9)2
=ln2/ln3+ln2/ln9
=ln2/ln3+ln2/(ln3^2)
=ln2/ln3+ln2/(2ln3)
=(ln2/ln3)(1+1/2)
=(ln2/ln3)*3/2
[log(4)3+log(8)3][log(3)2+log(9)2]
=(5/6)(ln3/ln2)*(ln2/ln3)*3/2
=5/4
log(1/2)32∧1/4
=(1/4)log(1/2)32
=(1/4)log(1/2)(2^5)
=(5/4)log(1/2) 2
=(5/4)log(1/2) (1/2)^(-1)
=-5/4
所以
[log(4)3+log(8)3][log(3)2+log(9)2]-log(1/2)32∧1/4
=5/4-(-5/4)
=5/4+5/4
=(5/4)*2
=5/2
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