Question

Se log 2 = a e log 3 = b, calcule em função de a e b o valor de log3√1,08

Answer 1

$\sqrt{a*b*c}=\sqrt{a}*\sqrt{b}*\sqrt{c}$
$\sqrt[n]{a^{x}}=a^{x/n}$

$log_{b}(a*c)=log_{b}(a)+log_{b}(c)$
$log_{b}(a^{n})=n*log_{b}(a)$
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$log_{3}\sqrt{1,08}=log_{3}\sqrt{108/100}$
$log_{3}\sqrt{1,08}=log_{3}[(1/10)\sqrt{108}]$
$log_{3}\sqrt{1,08}=log_{3}[(1/10)\sqrt{4*9*3}]$
$log_{3}\sqrt{1,08}=log_{3}[(1/10)*2*3* \sqrt{3}]$
$log_{3}\sqrt{1,08}=log_{3}[10^{-1}*2*3*3^{1/2}]$
$log_{3}\sqrt{1,08}=log(10^{-1}*2*3*3^{1/2})/log(3)$
$log_{3}\sqrt{1,08}=(log(10^{-1})+log(2)+log(3)+log(3)^{1/2})/b$
$log_{3}\sqrt{1,08}=[[-1]*log(10)+a+b+(1/2)*log(3)]/b$
$log_{3}\sqrt{1,08}=([-1]*1 + a + b + (1/2)b])/b$
$log_{3}\sqrt{1,08}=(- 1 + a + b + [b / 2]))/b$
$log_{3}\sqrt{1,08}=[(-2+2a+2b+b)/2]/b$
$log_{3}\sqrt{1,08}=(2a+3b-2)/2b$