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[alibaba8000]

1/A
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c) \(log_ \frac{54}{}168\), jeśli \(log_712=a\) i \(log_ \frac{12}{}24=b\)

[anka]

Rozwiązanie znalezione w necie:

\(log_{7}12=a \Rightarrow log_{12}7=\frac{1}{a}\)

\(log_{12}24=b\)
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\(log_{12}24=\frac{log_224}{log_212}=\frac{log_2(2^3\cdot3)}{log_2(2^2\cdot 3)}=\)

\(\frac{log_22^3+log_23}{log_22^2+log_23}=\frac{3log_22+log_23}{2log_22+log_23}=\frac{3+log_23}{2+log_23}\)

\(\frac{3+log_23}{2+log_23}=b\)

\(3+log_23=b(2+log_23)\)

\(3+log_23=2b+blog_23\)

\(log_23-blog_23=2b-3\)

\(log_23(1-b)=2b-3\)

\(log_23=\frac{2b-3}{1-b}\)
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\(log_{54}168=\frac{log_{12}168}{log_{12}54}=\frac{log_{12}(24\cdot 7)}{log_{12}54}=\frac{log_{12}24+log_{12}7}{log_{12}54}\)

Licznik:
\(log_{12}24+log_{12}7=b+\frac{1}{a}=\frac{ab+1}{a}\)

Mianownik:

\(log_{12}54=\frac{log_2 54}{log_212}=\frac{log_2(2\cdot 3^3)}{log_2(2^2\cdot 3)}=\frac{log_22+log_2 3^3}{log_22^2+log_23}\)

\(\frac{1+3log_2 3}{2log_22+log_23}=\frac{1+3log_2 3}{2+log_23}=\frac{1+3\cdot \frac{2b-3}{1-b}}{2+\frac{2b-3}{1-b}}=\)

\(\frac{\frac{1-b+3(2b-3)}{1-b}}{\frac{2(1-b)+2b-3}{1-b}}=\frac{1-b+3(2b-3)}{2(1-b)+2b-3}=\)

\(\frac{1-b+6b-9}{2-2b+2b-3}=\frac{5b-8}{-1}=8-5b\)
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\(log_{54}168=\frac{\frac{ab+1}{a}}{8-5b}=\frac{ab+1}{a(8-5b)}\)