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log9 27

log9 27 =log9 (9*3) =log9 9+log9 3 =1+log9 9^(1/2) =1+1/2 =3/2

换底公式:loga b/loga c=logc b 所以原式=log9 27 =log3² 3³ =(3/2)*log3 3 =3/2

额,要不你发图后面那个5开始就不知道你是啥式子,前面2个计算如图

log9 27=log9 (√9)³=log9 (9)^(3/2)=3/2。

可以用换底公式,换成你熟悉的底数的对数。例如:常用对数、自然对数。 log(a,b)=lgb/lga log(8,9 )* log(27,32) =lg9/lg8*lg32/lg27 =2lg3*5lg2/3lg2*3lg3 =10/9

只要使用公式 log(a^n底)(b^m))=m/n*loga b (log2 3+log4 9.+log8 27+.+log(2^n底)(3^n)这些相加的项都是相等的 log2^n(3^n)=log2(3) 还有就是log2 3=1/log3 2 (log2 3+log4 9+log8 27+……+log2^n 3^n)log9 n次根号32 =nlog2 (3)*1/n*5/2...

题干不全,缺少具体题干内容,无法作答。

y = log9[x/27] * log(1/3)[3x] = log3[√(x/27) * log3[1/(3x)] = 1/2{ log3[x]-log3[27] } * { -log3[3]-log3[x] } = -1/2{ log3[x]-3 } * {1 + log3[x] } = -1/2 { log3[x] } ² + log3[x] + 3/2 = -1/2 { log3[x] - 1 } ² + 1 x属于...

换底公式

log4 27*log5 8*log9 625 =3/2 log23*3log5 2*2log3 5 =9*lg3/(lg2)*lg2/(lg5)*lg5/(lg3) =9

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