Логаритъм

[Гост]

log12(18)= a, log8(9)=?

[Davids]

Ако означим $log_89 = x$, то ще имаме:
$a = log_{12}18 = \frac{log_218}{log_212} = \frac{log_2(2.9)}{log_212} = \frac{log_22 + log_29}{log_212} = \frac{1 + 3log_89}{log_212} = \frac{3x + 1}{log_212}$
$\Rightarrow 3x + 1 = alog_212$
$x = \frac{alog_212 - 1}{3}$

[S.B.]

log12(18)= a, log8(9)=?

[tex]a = log_{12 }18 = log_{12 }2.9 = log_{12 }2 + log_{12 }3^{2} = \frac{log_{2 }2}{log_{2 }12} + 2log_{12 }3 = \frac{1}{log_{2 }12} + \frac{2log_{2 }3}{log_{2 }12} =[/tex]
[tex]= \frac{1 + 2log_{2 }3}{log_{2 }12} = \frac{1 + 2log_{2 }3}{log_{2 }3.4} = \frac{1 + 2 log_{2 }3}{log_{2 }3 + log_{2 }2^{2}} = \frac{1 + 2log_{2 }3}{log_{2 }3 + 2log_{2 }2} = \frac{1 + 2log_{2 }3}{log_{2 }3 + 2}[/tex]
Или [tex]a = \frac{1 + 2log_{2 }3}{log_{2 }3 + 2}[/tex]
[tex]log_{8 }9 = log_{8 }3^{2} = 2log_{8 }3 = 2\frac{log_{2 }3}{log_{2 }8} = \frac{2log_{2 }3}{log_{2 }2^{3}} = \frac{2log_{2 }3}{3log_{2 }2} = \frac{2log_{2 }3}{3}[/tex]
Или [tex]log_{8 }9 = \frac{2}{3}log_{2 }3[/tex]
От [tex]а = \frac{1 + 2log_{2 }3}{log_{2 }3 + 2 } \Leftrightarrow a.log_{2 }3 + 2a = 1 + 2.log_{2 }3 \Leftrightarrow log_{2 }3(a - 2) = 1 - 2a \Rightarrow log_{2 }3 = \frac{1 - 2a}{a - 2}[/tex]
Окончателно от [tex]log_{8 }9 = \frac{2}{3}.log_{2 }3 \Rightarrow log_{8 }9 = \frac{2(1 - 2a)}{3(a - 2)}[/tex]