Question

Sendo logx 2=a, logx 3=b calcule logx 3√12

Answer 1

$log_{x} 2 = a$
$log_{x} 3 = b$
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$log_{x}(3 \sqrt{12}) = log_{x} (3 \sqrt{3*4})$
$log_{x}(3 \sqrt{12}) = log_{x} [3( \sqrt{3} * \sqrt{4})]$
$log_{x}(3 \sqrt{12})=log_{x} [3( \sqrt{3} *2)]$
$log_{x}(3 \sqrt{12})=log_{x}(3*2*\sqrt{3})$
$log_{x}(3 \sqrt{12})=log_{x}(3*2*3^{1/2})$
$log_{x}(3 \sqrt{12})= log_{x}3+log_{x}2 + log_{x} 3^{1/2}$
$log_{x}(3 \sqrt{12})= b + a + (1/2)*log_{x}3$
$log_{x}(3 \sqrt{12})= b + a + (1/2)*b$
$log_{x}(3 \sqrt{12})= b + a + (b/2)$

Multiplicando todos os membros por 2:

$2*log_{x}(3 \sqrt{12})=2*b+2*a+2*(b/2)$
$2*log_{x}(3 \sqrt{12})=2b + 2a + b$
$2*log_{x}(3 \sqrt{12})=2a + 3b$
$log_{x}(3 \sqrt{12})=(2a+3b)/2$