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  • Roma
  • Community Manager
2013-07-17T23:52:38+02:00

Skorzystamy ze wzoru na zamianę podstawy logarytmu:

log_a \ b = \frac{log_c \ b}{log_c \ a}

 

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log_2 \ 360 = \frac{log_3 \ 360}{log_3 \ 2} = \frac{log_3 \ (5 \cdot 9 \cdot 8)}{log_3 \ 2} = \frac{log_3 \ 5 + log_3 \ 9+ log_3 8}{log_3 \ 2} =\\\\ = \frac{log_3 \ 5 + log_3 \ 9}{log_3 \ 2} + \frac{log_3 8}{log_3 \ 2} = \frac{log_3 \ 5 + log_3 \ 3^2}{log_3 \ 2} + \frac{log_3 2^3}{log_3 \ 2} =\frac{log_3 \ 5 + 2 \cdot log_3 \ 3}{log_3 \ 2} + \\\\ + \frac{3 \cdot log_3 2}{log_3 \ 2} = \frac{log_3 \ 5 + 2 \cdot 1}{log_3 \ 2} + 3 \cdot 1 =\frac{log_3 \ 5 + 2}{log_3 \ 2} + 3

 

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log_3 \ 15 = b \\\ log_3 \ (3 \cdot 5) = b \\\ log_3 \ 3 + log_3 \ 5 = b \\\ 1 + log_3 \ 5 = b \\\ log_3 \ 5 = b - 1

 

log_3 \ 20 = a \\\ log_3 \ (4 \cdot 5) = a \\\ log_3 \ 4 + log_3 \ 5 = a \\\ log_3 \ 4 = a - log_3 \ 5 \\\ log_3 \ 2^2 = a - (b - 1) \\\ 2 \cdot log_3 \ 2 = a - b +1 \ /:2 \\\ log_3 \ 2 = \frac{a-b+ 1}{2}

 

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log_2 \ 360 =\frac{log_3 \ 5 + 2}{log_3 \ 2} + 3 = \frac{b - 1 + 2}{\frac{a-b+ 1}{2}} + 3 = (b + 1) : \frac{a-b+ 1}{2}+ 3 = \\\\ = (b +1) \cdot \frac{2}{a-b+ 1}+ 3 =\frac{2b+2}{a-b+ 1}+ 3 =\frac{2b+2}{a-b+ 1}+ \frac{3 \cdot (a-b+ 1)}{a-b+ 1} = \\\\ = \frac{2b+2}{a-b+ 1}+ \frac{3a-3b+ 3}{a-b+ 1} = \frac{2b+2+3a-3b+ 3}{a-b+ 1} =\frac{3a-b+ 5}{a-b+ 1}

 

 

Odp. log_2 \ 360 =\frac{3a-b+ 5}{a-b+ 1}

 

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