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a)
| a | α | b | \(a^{\alpha}\cdot a^{\beta}\) | \(a^{\alpha}:a^{\beta}\) | \(a^{\alpha+\beta}\) | \(\alpha^{\alpha+\beta}\) |
| 3 | \(\sqrt{2}\) | \(\sqrt{3}\) | \(3^{\sqrt{2}}\cdot3^{\sqrt{3}}=31,70659\) | \(3^{\sqrt{2}}:3^{\sqrt{3}}=0,70527\) | \(3^{\sqrt{2}+\sqrt{3}}=31,70659\) | \(3^{\sqrt{2}-\sqrt{3}}=0,70527\) |
b) Nhận xét:
\(a^m\cdot a^n=a^{m+n};a^m:a^n=a^{m-n}\)
a: \(log_{\dfrac{1}{4}}8=log_{2^{-2}}2^3=\dfrac{-3}{2}\cdot log_22=-\dfrac{3}{2}\)
b: \(log_45\cdot log_56\cdot log_68\)
\(=log_45\cdot\dfrac{log_46}{log_45}\cdot\dfrac{log_48}{log_46}\)
\(=log_48=log_{2^2}2^3=\dfrac{3}{2}\)
\(a,A=log_23\cdot log_34\cdot log_45\cdot log_56\cdot log_67\cdot log_78\\ =log_28\\ =log_22^3\\ =3\\ b,B=log_22\cdot log_24...log_22^n\\ =log_22\cdot log_22^2...log_22^n\\ =1\cdot2\cdot...\cdot n\\ =n!\)
a: \(log_49=\dfrac{log9}{log4}=\dfrac{log3^2}{log2^2}=\dfrac{2\cdot log3}{2\cdot log2}=\dfrac{log3}{log2}=\dfrac{b}{a}\)
b: \(log_612=\dfrac{log12}{log6}=\dfrac{log2^2+log3}{log2+log3}=\dfrac{2\cdot log2+log3}{log2+log3}\)
\(=\dfrac{2a+b}{a+b}\)
c: \(log_56=\dfrac{log6}{log5}=\dfrac{log\left(2\cdot3\right)}{log\left(\dfrac{10}{2}\right)}=\dfrac{log2+log3}{log10-log2}\)
\(=\dfrac{a+b}{1-a}\)
a: l o g 4 9 = l o g 9 l o g 4 = l o g 3 2 l o g 2 2 = 2 ⋅ l o g 3 2 ⋅ l o g 2 = l o g 3 l o g 2 = b a log 4 9= log4 log9 = log2 2 log3 2 = 2⋅log2 2⋅log3 = log2 log3 = a b b: l o g 6 12 = l o g 12 l o g 6 = l o g 2 2 + l o g 3 l o g 2 + l o g 3 = 2 ⋅ l o g 2 + l o g 3 l o g 2 + l o g 3 log 6 12= log6 log12 = log2+log3 log2 2 +log3 = log2+log3 2⋅log2+log3 = 2 a + b a + b = a+b 2a+b c: l o g 5 6 = l o g 6 l o g 5 = l o g ( 2 ⋅ 3 ) l o g ( 10 2 ) = l o g 2 + l o g 3 l o g 10 − l o g 2 log 5 6= log5 log6 = log( 2 10 ) log(2⋅3) = log10−log2 log2+log3 = a + b 1 − a = 1−a a+b
a) \(log_50,5=-0,439677\)
c) \(In\left(\dfrac{3}{2}\right)=0,405465\)