Дано [tex]log _{6} 15=a[/tex]; [tex]log _{12} 18=b[/tex] Найти [tex]log _{25} 24[/tex]. Created by Jameus123. algebra-ru

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Дано [tex]log _{6} 15=a[/tex]; [tex]log _{12} 18=b[/tex]
Найти [tex]log _{25} 24[/tex]

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Матов

Verified answer

$log_{6}15=a\\
log_{12}18=b\\\\
log_{25}24=\frac{log_{5}24}{2} = \frac{2log_{5}2+log_{5}6}{2}\\\\ 
log_{12}18=\frac{2log_{6}3+log_{6}2}{2log_{6}2+log_{6}3}=b\\
log_{6}15=log_{6}3+log_{6}5=a\\\\
log_{6}2=\frac{log_{5}2}{log_{5}6}\\\\
log_{12}18=\frac{2log_{6}3+\frac{log_{5}2}{log_{5}6}}{2*\frac{log_{5}2}{log_{5}6}+log_{6}3}=b\\
log_{6}15=log_{6}3+\frac{1}{log_{5}6} = a\\\\
\\
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откуда $><br />заменяя <img src=$
и решая систему

$\frac{y+1}{x+y}=a\\ 
 \frac{x+2y}{2x+y}=b$
подставляя
получим
$log_{25}{24}=\frac{5-b}{2ab+2a-4b+2}$

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