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\(B=\log_{25}15\) biết \(\log_{25}3=a\)
Ta có : \(a=\log_{15}3=\frac{1}{\log_3\left(3.5\right)}=\frac{1}{1+\log_35}\)
\(\Rightarrow\log_35=\frac{1}{a}-1=\frac{1-a}{a}\)
\(\Rightarrow B=\log_{25}15=\frac{\log_315}{\log_325}=\frac{\log_3\left(3.5\right)}{\log_35^2}=\frac{1+\frac{1-a}{a}}{2.\log_35}=\frac{1}{2\left(1-a\right)}\)
Ta có :
\(a=\log_{14}7=\frac{1}{\log_7\left(2.7\right)}=\frac{1}{1+\log_72}\Rightarrow\log_72=\frac{1}{a}-1=\frac{1-1}{a}\)
\(b=\log_{15}5=\frac{\log_75}{\log_7\left(7.2\right)}=\frac{\log_72}{1+\log_72}\Rightarrow\log_75=b\left(1+\log_72\right)=b\left(1+\frac{1-a}{a}\right)=\frac{b}{a}\)
\(\Rightarrow E=\log_{35}28=\frac{\log_727}{\log_735}=\frac{\log_7\left(7.2^2\right)}{\log_7\left(7.5\right)}=\frac{1+\log_72}{1+\log_75}=\frac{1+2.\frac{1-a}{a}}{1+\frac{b}{a}}=\frac{2-a}{a+b}\)
a) \(A=\log_{5^{-2}}5^{\frac{5}{4}}=-\frac{1}{2}.\frac{5}{4}.\log_55=-\frac{5}{8}\)
b) \(B=9^{\frac{1}{2}\log_22-2\log_{27}3}=3^{\log_32-\frac{3}{4}\log_33}=\frac{2}{3^{\frac{3}{4}}}=\frac{2}{3\sqrt[3]{3}}\)
c) \(C=\log_3\log_29=\log_3\log_22^3=\log_33=1\)
d) Ta có \(D=\log_{\frac{1}{3}}6^2-\log_{\frac{1}{3}}400^{\frac{1}{2}}+\log_{\frac{1}{3}}\left(\sqrt[3]{45}\right)\)
\(=\log_{\frac{1}{3}}36-\log_{\frac{1}{3}}20+\log_{\frac{1}{3}}45\)
\(=\log_{\frac{1}{3}}\frac{36.45}{20}=\log_{3^{-1}}81=-\log_33^4=-4\)
1.
\(A=3log_{2^2}\sqrt{a}-log_{2^{-1}}a^2+2log_{a^{\dfrac{1}{2}}}a\)
\(=3.\dfrac{1}{2}.\dfrac{1}{2}log_2a-\left(-1\right).2.log_2a+2.2.log_2a\)
\(=\dfrac{27}{4}log_2a\)
2.
\(log_{12}36=\dfrac{log_236}{log_212}=\dfrac{log_2\left(3^2.2^2\right)}{log_2\left(3.2^2\right)}=\dfrac{log_23^2+log_22^2}{log_23+log_22^2}\)
\(=\dfrac{2.log_23+2}{log_23+2}=\dfrac{2a+2}{a+2}\)
cho e hỏi tại sao \(3\log_{2^2}\sqrt{a}\) lại bằng \(3.\dfrac{1}{2}.\dfrac{1}{2}\log_2a\) và \(2\log_{a^{\dfrac{1}{2}}}a=2.2.\log_2a\)
\(log_{c+b}a+log_{c-b}a=\frac{1}{log_a\left(c+b\right)}+\frac{1}{log_a\left(c-b\right)}\)
\(=\frac{log_a\left(c-b\right)+log_a\left(c+b\right)}{log_a\left(c-b\right).log_a\left(c+b\right)}=\frac{log_a\left(c^2-b^2\right)}{log_a\left(c-b\right)log_a\left(c+b\right)}\)
\(=log_aa^2.log_{\left(c+b\right)}a.log_{c-b}a=2log_{c+b}a.log_{c-b}a\)
a) Ta có 1350 = 30.32 . 5 suy ra
log301350 = log30(30. 32. 5) = 1 + 2log303 + log305 = 1 + 2a + b.
b) log2515 =
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