Given that a^2+b^2=12ab,where a >b.Prove log(a-b)=1/2(1+log a+log b)
Help!!!!!!!!!!!
math logarithmic
2014-07-02 9:16 pm
回答 (1)
2014-07-02 10:10 pm
✔ 最佳答案
(a - b)^2 = a^2 + b^2 - 2ab= 12ab - 2ab = 10ab
so log(a - b)^2 = log(10ab)
= 2 log ( a - b) = log 10 + log a + log b
so log (a - b) = (1 + log a + log b)/2
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