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Nhận xét : \(lg\tan1^0+lg\tan89^0=lg\left(\tan1^0.\tan89^0\right)=lg1=0\)
\(lg\tan2^0+lg\tan88^0=lg\left(\tan1^0.\tan88^0\right)=lg1=0\)
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Và \(lg\tan45^0=lg1=0\)
Suy ra \(S=lg\tan1^0+lg\tan2^0+lg\tan3^0+......+lg\tan89^0\)
\(=\left(lg\tan1^0+lg\tan89^0\right)+\left(lg\tan2^0+lg\tan88^0\right)+....+lg\tan45^0\)
Vậy \(S=lg\tan1^0+lg\tan2^0+lg\tan3^0+...+lg\tan89^0=0\)
\(G=lg\left(25^{\log_56}+49^{\log_78}\right)-e^{\ln3}=lg\left[\left(5^2\right)^{\log_56}+\left(7^2\right)^{\log_78}\right]-3\)
\(=lg\left(5^{\log_56^2}+7^{\log_78^2}\right)-3\)
\(=lg\left(6^2+8^2\right)-3=lg10^{2-3}=2-3=-1\)
a) Sử dụng công thức \(\frac{1}{\log_ba}=\log_ab\), hơn nữa \(x=2007!\) nên ta có : \(A=\log_x2+\log_x3+..........\log_x2007\)
\(=\log_x\left(2.3...2007\right)\)
\(=\log_xx=1\)
b) Nhận thấy
\(lg\tan1^o+lg\tan89^o=lg\left(lg\tan1^o.lg\tan89^o\right)=lg1=0\)
Tương tự ta có :
\(lg\tan2^o+lg\tan88^o=0\)
.................
\(lg\tan44^o+lg\tan46^o=0\)
\(lg\tan45^o=lg1=0\)
Do đó :
\(B=\left(lg\tan1^o+lg\tan89^o\right)+\left(lg\tan2^o+lg\tan88^o\right)+......+lg\tan45^0=0\)
\(I=lg\left(\sqrt{81^{\log_35}+27^{\log_936}}+3^{2\log_971}\right)=lg\left(\sqrt{\left(3^4\right)^{\log_35}+\left(3^3\right)^{\log_{3^2}6^2}}+3^{2\log_{3^2}71}\right)\)
\(=lg\left(\sqrt{3^{\log_35^4}+3^{\log_36^3}}+3^{\log_371}\right)=lg\left(\sqrt{5^4+6^3}+71\right)\)
\(=lg\left(29+71\right)=lg100=2\)
câu b
<=> lg(2x+4) = lg(|4x-7|)2
<=> 2x+4 = 16x2- 56x + 49 <=> x=2,5 hoặc x= 1,125
Theo giả thiết ta có : \(x^2+4y^2=12xy\Leftrightarrow\left(x+2y\right)^2=16xy\)
Do \(x,y>0\Rightarrow x+2y=4\sqrt{xy}\)
Khi đó ta có :
\(lg\left(x+2y\right)=lg4+\frac{1}{2}lgxy\Leftrightarrow lg\left(x+2y\right)-2lg2=\frac{1}{2}\left(lgx+lgy\right)\)
Vậy với \(x,y>0\) và \(x^2+4y^2=12xy\) thì \(lg\left(x+2y\right)-2lg2=\frac{1}{2}\left(lgx+lgy\right)\)
\(M=lg\left|\log_{\frac{1}{a^3}}\sqrt[5]{a\sqrt{a}}\right|=lg\left|\log_{\frac{1}{a^3}}\sqrt[5]{a.a^{\frac{1}{2}}}\right|=lg\left|\log_{\frac{1}{a^3}}\left(a^{\frac{3}{2}}\right)^{\frac{1}{5}}\right|=lg\left|\log_{a^{-3}}a^{\frac{3}{10}}\right|=lg\left|-\frac{1}{10}=lg\frac{1}{10}=-1\right|\)
\(N=lg\left(\tan1^0\right)+lg\left(\tan2^0\right)+....+lg\left(\tan88^0\right)+lg\left(\tan89^0\right)\)
\(=\left[lg\left(\tan1^0\right)+lg\left(\tan89^0\right)\right]+\left[lg\left(\tan2^0\right)+lg\left(\tan88^0\right)\right]+...+\left[lg\left(\tan44^0\right)+lg\left(\tan46^0\right)\right]+lg\left(\tan45^0\right)\)
\(=lg\left(\tan1^0.\tan89^0\right)+lg\left(\tan2^0.\tan88^0\right)+...+lg\left(\tan44^0.\tan46^0\right)+lg\left(\tan45^0\right)\)
\(=lg\left(\tan1^0.\cot1^0\right)+lg\left(\tan2^0.\cot2^0\right)+.....+lg\left(\tan44^0.\cot44^0\right)+lg\left(\tan45^0\right)\)
\(=lg1+lg1+....+lg1+lg1=0+0+....+0+0=0\)