已知log8`9=a,log9`25=b,用a.b表示lg2
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log89=lg9/lg8=2lg3/3lg2=a;log320=lg20/lg3=(1+lg2)/lg3=b;从此建立二元一次方程组,解得lg2=2/(3ab-2),
用换底公式.原式=(lg3/lg2+2lg3/3lg2)*(2lg2/lg3+3lg2/2lg3+lg2/lg3)=15/2
公式loga²(b)=1/2loga(b) loga(b)×logb(a)=1
a=log4(3)=lg3/lg4所以lg3=alg4=2alg2log8(9)=lg9/lg8=2lg3/3lg2=4alg2/3lg2=4a/3log4(3)=lg3/2lg2=(1/2)*log
∵a,b,c都是正实数,且满足log9(9a+b)=log3ab,∴log9(9a+b)=log3ab=log9ab,∴9a+b=ab,∴9a+bab=9b+1a=1,∴4a+b=(4a+b)(9b+
要用到换底公式,以及对数运算法则如log49=2/2*log₂3=log₂3log(1/3)⁴√81=-log₃⁴√81(log49+log
(log23/log24+log23/log28)(log35/log33+log35/log39)(log52/log55+log52/log525)=[(5log23)/6][(3log35)/2
(log4^3+log8^9)*(log3^2+log9^16)=(log2^3/log2^4+log2^9/log2^8)*(log3^2+log3^16/log3^9)=(1/2log2^3+1/
注明一下loga的n次方b的m次方=(m/n)logablogab=lgb/lga原式=(log23+log89)(log34+log98+log32)=[log23+(2/3)log23][2log
tanA+tanB=2(log872+log972)=2(log89+log88+log98+log99)=2[(2/3)log23+(3/2)log32+2]所以(sinAcosB+cosAsinB
a=log89=lg9/lg8=2lg3/3lg2b=log320=lg20/lg3=(1+lg2)/lg3ab=2(1+lg2)/3lg2=(2/3lg2)+1/3ab-1/3=2/3lg2lg2=
log92/3由于3>根号8因此log8根号3>log8根号下(根号8)=1/4因此a
log95=log75/log79=a(换底公式)又log79=b则log75=ab同理log359=log79/log735log735=log75+log77=ab+1所以log359=b/(ab
由log95=a,log97=b得log935=log95+log97=a+b所以log359=1/log935=1/(a+b)
log8^3=a/3,log8^9=2a/3
先介绍换底公式:log(a)b=lgb/lga证明:设log(a)b=t则a^t=b,两边取以10为底的对数lga^t=lgbtlga=lgb所以t=lgb/lga所以log(a)b=lgb/lga=
因为loga(b)=lgb/lga所以原式可化为lg10/lg5=log5(10)
(1)(2)(3)怎么用这个范围太大了,只能说因题而异再问:lg18等于log18吗再答:有的计算器默认log是lg再问:知道了谢谢再答:不客气