2024-2025学年新教材高中数学课时素养检测三十一第四章指数函数与对数函数4.3.2对数的运算含解析新人教A版必修第一册

PAGE课时素养检测三十一对数的运算(30分钟60分)一、选择题(每小题5分,共30分)1.计算QUOTE-QUOTE的值为 ()A.-QUOTE B.-QUOTE C.2 D.1【解析】选D.QUOTE-QUOTE=QUOTE-QUOTE=QUOTE-QUOTE=1.2.已知2m=3n=36,则QUOTE+QUOTE= ()A.1 B.2 C.QUOTE D.QUOTE【解析】选D.因为2m=3n=36,则m=log236,n=log336,所以QUOTE+QUOTE=log362+log363=log36QUOTE=log366=QUOTE=QUOTE.3.设f(x)=logax(a>0,且a≠1),对于随意的正实数x,y,都有 ()A.f(xy)=f(x)f(y) B.f(xy)=f(x)+f(y)C.f(x+y)=f(x)f(y) D.f(x+y)=f(x)+f(y)【解析】选B.由对数运算法则f(xy)=loga(xy)=logax+logay,所以f(xy)=f(x)+f(y).4.设函数f(x)=logax(a>0且a≠1),若loga(x1x2…x2020)=7,则f(QUOTE)+f(QUOTE)+…+f(QUOTE)= ()A.8 B.21 C.16 D.24【解析】选B.f(QUOTE)+f(QUOTE)+…+f(QUOTE)=3f(x1)+3f(x2)+…+3f(x2020)=3loga(x1x2…x2020)=21.5.设2a=7b=m且QUOTE+QUOTE=2,则m= ()A.2QUOTE B.QUOTE C.±QUOTE D.±2QUOTE【解析】选B.2a=7b=m,所以a=log2m,b=log7m,QUOTE+QUOTE=logm2+logm7=logm14=2,所以m=QUOTE.6.若5a=2b=1QUOTE且abc≠0,则QUOTE+QUOTE= ()A.1 B.2 C.3 D.4【解析】选B.设5a=2b=1QUOTE=t>0,则a=log5t,b=log2t,QUOTE=lgt,QUOTE+QUOTE=QUOTE+QUOTE=2lgt(logt5+logt2)=2lgtlogt10=2.二、填空题(每小题5分,共10分)7.计算(lg5)2-(lg2)2+2lg2=________.

【解析】(lg5)2-(lg2)2+2lg2=(lg5+lg2)(lg5-lg2)+2lg2=lg5-lg2+2lg2=lg5+lg2=1.答案:18.3log72-log79+2log7QUOTE=________.

【解析】原式=log723-log79+log7QUOTE=log7QUOTE=log71=0.答案:0三、解答题(每小题10分,共20分)9.计算:(1)2log63+log64.(2)QUOTE÷10QUOTE.(3)log43·log98.(4)log2.56.25+lnQUOTE-0.06QUOTE.【解析】(1)原式=log632+log64=log6(32×4)=log6(62)=2log66=2.(2)原式=lgQUOTE÷1QUOTE=lg102÷10-1=2×10=20.(3)原式=QUOTE·QUOTE=QUOTE·QUOTE=QUOTE.(4)原式=log2.5(2.5)2+QUOTE-QUOTE=2+QUOTE-QUOTE=QUOTE.10.已知loga(x2+4)+loga(y2+1)=loga5+loga(2xy-1)(a>0且a≠1),求log8QUOTE的值.【解析】由对数的运算法则,可将等式化为loga[(x2+4)·(y2+1)]=loga[5(2xy-1)],所以(x2+4)(y2+1)=5(2xy-1).整理,得x2y2+x2+4y2-10xy+9=0,配方,得(x