Se o logaritmo de 4 na base 6 é igual a "a", então o logaritmo de 32 na base 9R: 5a/ 4-2a.... Pergunta de ideia dejaruaba20

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Se o logaritmo de 4 na base 6 é igual a "a", então o logaritmo de 32 na base 9

R: 5a/ 4-2a

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lamacch $log_{6} 4=a$ ⇒ $log_{6} 2^{2} =a$ ⇒ $2.log_{6} 2 =a$ ⇒ $log_{6} 2 = \frac{a}{2}$ ⇒ $\frac{log_{3} 2}{log_{3} 6} =$ $\frac{log_{3} 2}{log_{3} 2.3} =\frac{log_{3} 2}{log_{3} 2+log_{3} 3}=\frac{log_{3} 2}{log_{3} 2+1}= \frac{a}{2}$

Façamos $log_{3} 2=x$ ⇒ $\frac{x}{x+1}= \frac{a}{2}$ ⇒ $2x=ax+a$ ⇒ $2x-ax=a$ ⇒ $(2-a).x=a$ ⇒ $x= \frac{a}{2-a}$

Logo: $log_{3} 2=\frac{a}{2-a}$

$log_{9} 32=log_{9} 2^{5} =5log_{9} 2=$

$=5. \frac{log_{3} 2}{log_{3} 3^{2} } =5. \frac{log_{3} 2}{2.log_{3} 3 } =5. \frac{log_{3} 2}{2.1 } = \frac{5}{2} .log_{3} 2=\frac{5}{2} .(\frac{a}{2-a})=\frac{5a}{4-2a}$

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