tex]...

May 2021 2 30 Report

решите логарифм
[tex]a) log_{3} 63 - log_{3} 9 + 0.5log_{3} \frac {27}{49} [/tex]
[tex]b) log_{2} 27 - 2log_{2} 3 + log_{2} \frac {2}{3} [/tex]

xti $log_{3} 63- log_{3} 9+ log_{3} \sqrt{ \frac{27}{49}}= log_{3} ( \frac{63}{9} * \frac{3 \sqrt{3} }{7} )= log_{3}3^{ \frac{3}{2} } } =1.5$

$log_{2} 27-log_{2} 9+ log_{2} \frac{2}{3} =log_{2} ( \frac{27}{9} * \frac{2}{3} )= log_{2} 2=1$

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mukus13

Verified answer

A)
$log_3(63)-log_3(9)+0.5*log_3(27/49) =$
$log_3(63)-2+0.5*(log_3(27) - log_3(49))=$
$log_3(63) -2 + 0.5*3 - 0.5*log_3(49)=$
$log_3(63) -2 + 0.5*3 - log_3(49^{1/2} )=$
$log_3(63) -2 + 0.5*3 -log_3(7)=$
$log_3(63/7) -2 + 0.5*3 =$
$log_3(9)-2+1.5=$
$2-2+1.5 = 1.5$
b)
$log_2(27)-2log_2(3)+log_2(2/3)=$
$log_2(27)-log_2(3^2)+log_2(2)-log_2(3)=$
$log_2(27)-log_2(3^2)-log_2(3)+1=$
$log_2(3^3)-( log_2(3^2) + log_2 (3) ) +1 =$
$log_2 (3^3) - log_2(3^2*3^1)+1=$
$log_2( \frac{3^3}{3^3} )+1 = 0+1=1$

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