已知a log32,b=log2 三分之一

log23=a;log37=b即lg3/lg2=a;lg7/lg3=b;所以lg7/lg2=a*b所以log72=1/a*b同理log142=1/1+ab;所以结果为（3+ab）/(1+ab)

lg2=a,lg3=b,log215=lg15/lg2=lg(3*5)/lg2=(lg3+lg5)/lg2=(lg3+lg(10/2))/lg2=(lg3+lg10-lg2)/lg2=(b+1-a)/

由题意知,x^2-2>0,解得x∈（-∞,负根号2）∪（正根号2,+∞）A.当x∈（-∞,负根号2）时,f(x)是单调递减函数.当f(x)=1时,x取得最大值,解log2^(x^2-2)＝1得到x=-

运用换底:log14(56)=log3(56)/log3(14)=〔log3(7)＋log3(8)〕/〔log3(7)＋log3(2)〕log3(2)=1/log2(3)=1/alog3(8)=3lo

log14^56=log3^56/log3^14=(log3^7+3log3^2)/(log3^7+log3^2)=(b+3/a)/(b+1/a)=(3+ab)/(1+a)

x^2-2>0|x|>√2x^2-2=2x=±2x^2-2=14x=±4∴a1=2,且b1=4a2=-4,且b2=-2.

运用换底:log14(56)=log3(56)/log3(14)=〔log3(7)＋log3(8)〕/〔log3(7)＋log3(2)〕log3(2)=1/log2(3)=1/alog3(8)=3lo

原式=lg15/lg2=[lg(3×5)]/a=(lg3+lg5)/a=[b+(1-lg2)]/a=(b-a+1)/a

log1256=lg56/lg12=(3lg2+lg7)/(lg3+lg4)因为a=lg3/lg2所以lg3=alg2因为b=lg7/lg3所以lg7=blg3=ablg2所以原式=(3lg2+abl

a=lg3/lg2b=lg7/lg3log4256=lg56/lg42=(lg7+3lg2)/(lg2+lg3+lg7)=(lg7/lg3+3lg2/lg3)/(lg2/lg3+1+lg7/lg3)=

log(2)3=a==>lg3/lg2=a==>lg3=alg2log(3)5=b==>lg5/lg3=b==>lg5=blg3=ablg2log(15)20=lg20/lg15=(lg2+lg2+l

log2(3)=a,则log3(2)=1/a.则log36(45)=[log3(36)]/[log3(45)]=[log3(4)＋log3(9)]/[log3(9)＋log3(5)]=[2log3(2

a=lg3/lg2,lg2=lg3/ab=lg5/lg3,lg5=blg3log15(20)=lg20/lg15=lg(2²*5)/lg(3*5)=(2lg2+lg5)/(lg3+lg5)=

f(x)=log2^(x^2-2):该函数定义域为全体实数,log对数是以10为底的对数,所以是增函数,而x^2-2的增区间为（0,正无穷）,减区间为（负无穷,0）,所以整个函数f(x)=log2^(

你好a=log[2]3√3＞1b=log[2]（9/√3）=log[2]3√3=ac=log[3]2＜1所以c＜b=a很高兴为您解答,祝你学习进步!有不明白的可以追问!如果有其他问题请另发或点击向我求

log2(bx)*log2(ax)+1=0(log2b+log2x)*(log2a+log2x)+1=0(log2x)²+(log2a+log2b)log2x+log2a*log2b+1=0

用换底公式：loga(b)=lgb/lgalg2=alg3=blog2(3)=lg3/lg2=b/a

log2(bx)×log2(ax)+1=0（log2b+log2x）（log2a+log2x）+1=0（log2x）2+（log2a+log2b）logx+log2alog2b+1=0（log2a+l

log2(bx)×log2(ax)+1=0（log2b+log2x）（log2a+log2x）+1=0（log2x）2+（log2a+log2b）logx+log2alog2b+1=0（log2a+l