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log2 4+log3 9=

log2+log3=log6 如下图所示,是“4”的第一个公式的逆运算

log2 4+log3 9=2+2=4.

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

loga(b)*logb(c)=loga(c); loga(b)*logb(c)*logc(d)=loga(d). log2(3)*log3(4)*log4(5)*log5(6)*log6(7)*log7(8)=log2(8)=3,所以3=loga(b),所以b=a^3

(log2 3+log4 9)(log3 4+log9 2) =[log(2)3+4log(2)3][2log(3)2+2log(3)2] =5log(2)3x4log(3)2 =20

原式=(lg3/lg4+lg3/lg8)(lg2/lg3+lg2/lg9)-log2^4-log2^(√32)=(lg3/2lg2+lg3/3lg2)(lg2/lg3+lg2/2lg3)-2-5/2=(1/2+1/3)(lg3/lg2)(1+1/2)(lg2/lg3)-2-5/2=(5/6)×(3/2)-2-5/2=5/4-2-5/2=-13/4

都换成以2为底的对数进行比较 log2(3)=log2(根号9) log4(5)=log2(根号5) 3/2=log2(2^(3/2))=log2(根号8) 因为y=log2(x)在(0,+无穷)上为增函数 所以log2(3)>3/2>log4(5)

换底公式轻松解决: 原式= (log3/log4+log3/log8)(log2/log3+log2/log9) (log是10为底) =(log3/(2log2)+log3/(3log2))(log2/log3+log2/(2log3)) =5/6*log3/log2*3/2*log2/log3 =5/4

答: 换底公式应用 log4(9)-log2(12) =log2(9)/log2(4)-log2(12) =0.5×2×log2(3)-log2(12) =log2(3)-log2(12) =log2(3/12) =log2(1/4) =-2

log (x-2) = log(x+4) log (x-2) = log(x+4) /log9 log (x-2)^2 = log(x+4) (x-2)^2 =x+4 x^2-5x=0 x= 5 or 0 (rej) ie x=5

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