Question

(IME) Considere o sistema de equações dado por
{3log3a + log9b = 10
{log9a - 2log3b = 10
onde a e b são números reais positivos.
Determine o produto P = a . b​

Answer 1

• O valor de P será igual a 1.

Para resolver sua questão, devemos aplicar as propriedades dos logs. Sendo elas:

$\large\displaystyle\begin{aligned} \log _a (b\cdot c) = \log _a b + \log_a c\end{aligned}$

$\large\displaystyle\begin{aligned} \log _a \left(\frac{b}{c} \right) = \log _a b - \log _a c \end{aligned}$

$\large\displaystyle\begin{aligned} \log _a (b^c) = c\cdot \log _a b \end{aligned}$

$\large\displaystyle\begin{aligned} \log _{a^c} (b) = \frac{1}{c} \cdot \log _a b \end{aligned}$

Dado então o sistema: $\large\displaystyle\begin{aligned} \begin{cases} 3\log _3a+ \log _9 b = 10\\ \log_9 a - 2\log _3b = 10\end{cases} \end{aligned}$ , como são equações logaritmas, temos que:

$\large\displaystyle\begin{aligned} 3\log _3a+ \log _9 b = 10 \end{aligned}$

$\large\displaystyle\begin{aligned} 3\log _3a+ \log _{3^2} b = 10 \end{aligned}$

$\large\displaystyle\begin{aligned} 3\log _3a+ \frac{1}{2} \log _{3} b = 10 \end{aligned}$

$\large\displaystyle\begin{aligned} 6\log _3a+ \log _{3} b = 20 \end{aligned}$

$\large\displaystyle\begin{aligned} \log _3a^6+ \log _{3} b = 20 \end{aligned}$

$\large\displaystyle\begin{aligned} \log _3(a^6\cdot b) = 20\ (I) \end{aligned}$

• Arrumando a equação II:

$\large\displaystyle\begin{aligned} \log_9 a - 2\log _3b = 10 \end{aligned}$

$\large\displaystyle\begin{aligned} \log_{3^2} a - 2\log _3b = 10 \end{aligned}$

$\large\displaystyle\begin{aligned} \frac{1}{2} \log_{3} a - 2\log _3b = 10 \end{aligned}$

$\large\displaystyle\begin{aligned} \log_{3} a - 4\log _3b = 20 \end{aligned}$

$\large\displaystyle\begin{aligned} \log_{3} a - \log _3b^4 = 20 \end{aligned}$

$\large\displaystyle\begin{aligned} \log_{3}\left( \frac{a}{b^4} \right)= 20 \ \ (II)\end{aligned}$

Pela definição de log, temos que: $\large\displaystyle\begin{aligned} \log _a (b) = x\end{aligned}$ , onde a^{x} tem que ser igual a b. Logo:

$\large\displaystyle\begin{aligned} \begin{cases}\log _3(a^6 \cdot b) = 20\ \ (I)\\ \log_3 \left(\frac{a}{b^4}\right) = 20\ \ (II)\end{cases} \end{aligned}$

$\large\displaystyle\begin{aligned} \log _3(a^6\cdot b) = 20\ \ (I) \end{aligned}$

$\large\displaystyle\begin{aligned} (a^6\cdot b) = 3^{20}\ \ (I) \end{aligned}$

$\large\displaystyle\begin{aligned} \log_{3}\left( \frac{a}{b^4} \right)= 20 \ \ (II)\end{aligned}$

$\large\displaystyle\begin{aligned} \left( \frac{a}{b^4} \right)= 3^{20} \ \ (II)\end{aligned}$

Então temos que aquele sistema cabuloso nada mais é doq esse sisteminha:

$\large\displaystyle\begin{aligned} \begin{cases}(a^6 \cdot b) = 3^{20}\ \ (I)\\ \left(\frac{a}{b^4}\right) = 3^{20}\ \ (II)\end{cases} \end{aligned}$

$\large\displaystyle\begin{aligned} \left( \frac{a}{b^4} \right)= 3^{20} \end{aligned}$

$\large\displaystyle\begin{aligned} \left( a \right)= b^4 \cdot 3^{20} \end{aligned}$

$\large\displaystyle\begin{aligned} (a^6\cdot b) = 3^{20} \end{aligned}$

$\large\displaystyle\begin{aligned} ((b^4\cdot3^{20})^6\cdot b) = 3^{20} \end{aligned}$

$\large\displaystyle\begin{aligned} b^{25}\cdot3^{120} = 3^{20} \end{aligned}$

$\large\displaystyle\text{ \begin{aligned} b^{25} = \frac{3^{20}}{3^{120} } \end{aligned} }$

$\large\displaystyle\begin{aligned} b^{25} = 3^{-100} \end{aligned}$

$\large\displaystyle\text{ \begin{aligned} b = \sqrt[25]{\frac{1}{3^{100}}} \end{aligned} }$

$\large\displaystyle\text{ \begin{aligned}\therefore \boxed{\boxed{\green{ b = \frac{1}{3^4} }}}\end{aligned} }$

Substitua na segunda equação.

$\large\displaystyle\begin{aligned} \left( a \right)= b^4 \cdot 3^{20} \end{aligned}$

$\large\displaystyle\begin{aligned} a = \left( \frac{1}{3^4} \right)^4 \cdot 3^{20} \end{aligned}$

$\large\displaystyle\begin{aligned} a= \left( 3^{-4} \right)^4 \cdot 3^{20} \end{aligned}$

$\large\displaystyle\begin{aligned} a= 3^{-16} \cdot 3^{20} \end{aligned}$

$\large\displaystyle\text{ \begin{aligned} \therefore \boxed{\boxed{\grenn{a= 3^{4}}}}\end{aligned} }$

• Logo, p = a * b . Sendo a = 3^{4} e b = (1/3^4).

$\large\displaystyle\begin{aligned} P= a\cdot b\end{aligned}$

$\large\displaystyle\text{ \begin{aligned} P= \not{3^{4}}\cdot \left( \frac{1}{\not{3^4}} \right)\end{aligned} }$

$\large\displaystyle\text{ \begin{aligned} \therefore\boxed{\boxed{\green{ P= 1}}}\ \checkmark\end{aligned} }$

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Answer 2

Resolvendo o sistema de equações logarítmicas encontramos que a = 3⁴ e b = 1/3⁴, assim o produto P = a · b é igual a 1.

⠀

Desejamos encontrar as incógnitas ‘‘a’’ e ’‘b’’ do seguinte sistema de equações para assim determinarmos o produto P = a · b

                      $\\\Large\boldsymbol{\begin{array}{l}\begin{cases}3\ell og\:\!_3\,(a)+\ell og\:\!_9\,(b)=10~~~\mathnormal{(\,I\,)}\\\\\ell og\:\!_9\,(a)-2\ell og\:\!_3\,(b)=10~~\mathnormal{(\,II\,)}\end{cases}\end{array}}$

, como são equações logarítmicas vamos usar algumas propriedades dos logaritmos, deixarei  listadas as que vamos usar aqui no início:

$\\\begin{array}{l}1^\circ)~~\ell og\:\!_a\,(b^c)~\Leftrightarrow~c\cdot\ell og\:\!_a\,(b)\\\\2^\circ)~~\ell og\:\!_{a^{c}}\,(b)~\Leftrightarrow~\dfrac{1}{c}\cdot\ell og\:\!_a\,(b)\\\\3^\circ)~~\ell og\:\!_a\,(b)+\ell og\:\!_a\,(c)~\Leftrightarrow~\ell og\:\!_a\,(b\cdot c)\\\\4^\circ)~~\ell og\:\!_a\,(b)-\ell og\:\!_a\,(c)~\Leftrightarrow~\ell og\:\!_a\,\bigg(\dfrac{b}{c}\bigg)\end{array}$

Dessa forma, vamos prosseguir.

⠀

Inicialmente, pensamos em deixar as bases dos logaritmos iguais em cada eq. para que possamos efetuar as propriedade 3° e 4°. Então,

⠀

na eq. ( ɪ ):

$\large\begin{array}{l}3\ell og\:\!_3\,(a)+\ell og\:\!_9\,(b)=10\\\\3\ell og\:\!_3\,(a)+\ell og\:\!_{3^2}\,(b)=10\\\\3\ell og\:\!_3\,(a)+\dfrac{1}{2}\cdot\ell og\:\!_3\,(b)=10\\\\2\cdot3\ell og\:\!_3\,(a)+\ell og\:\!_3\,(b)=2\cdot10\\\\6\ell og\:\!_3\,(a)+\ell og\:\!_3\,(b)=20\\\\\ell og\:\!_3\,(a^6)+\ell og\:\!_3\,(b)=20\\\\\ell og\:\!_3\,(a^6\cdot b)=20~~~\mathnormal{(\,III\,)}\end{array}$

⠀

; na eq. ( ɪɪ ):

$\large\begin{array}{l}\ell og\:\!_9\,(a)-2\ell og\:\!_3\,(b)=10\\\\\ell og\:\!_{3^2}\,(a)-2\ell og\:\!_3\,(b)=10\\\\\dfrac{1}{2}\cdot\ell og\:\!_3\,(a)-2\ell og\:\!_3\,(b)=10\\\\\ell og\:\!_3\,(a)+2\cdot[-2\ell og\:\!_3\,(b)]=2\cdot10\\\\\ell og\:\!_3\,(a)-4\ell og\:\!_3\,(b)=20\\\\\ell og\:\!_3\,(a)-\ell og\:\!_3\,(b^4)=20\\\\\ell og\:\!_3\,\bigg(\dfrac{a}{b^4} \bigg)=20~~~\mathnormal{(\,IV\,)}\end{array}$

Agora vamos formar um novo sistema com as equações ( ɪɪɪ ) e ( ɪᴠ ):

                                     $\\\Large\begin{array}{l}\begin{cases}\ell og\:\!_3\,(a^6\cdot b)=20~~~\mathnormal{(\,III\,)}\\\\\ell og\:\!_3\,\bigg(\dfrac{a}{b^4}\bigg)=20~~~\mathnormal{(\,IV\,)}\end{cases}\end{array}$

Veja que, pela definição de logaritmo temos que $\ell og\:\!_a\,(b)=c~\Leftrightarrow~b=a^c$, assim:

$\\\large\begin{array}{l}\implies~~\begin{cases}\ell og\:\!_3\,(a^6\cdot b)=20\\\\\ell og\:\!_3\,\bigg(\dfrac{a}{b^4}\bigg)=20\end{cases}\\\\\iff~~\begin{cases}a^6\cdot b=3^{20}\\\\\dfrac{a}{b^4}=3^{20}\end{cases}\end{array}$

Agora, usando o método da substituição, a fim de encontrar ‘‘b’’ podemos isolar ‘‘a’’ na segunda eq.:

$\\\large\begin{array}{l}\implies~~~\dfrac{a}{b^4}=3^{20}\\\\\iff~~~a=b^4\cdot3^{20}\end{array}$

E assim podemos substituí-lo na primeira eq.:

$\\\large\begin{array}{l}\implies~~~a^6\cdot b=3^{20}\\\\\iff~~~\big(b^4\cdot3^{20}\big)^6\cdot b=3^{20}\\\\\iff~~~(b^{24}\cdot3^{120})\cdot b=3^{20}\\\\\iff~~~b^{24+1}\cdot3^{120}=3^{20}\\\\\iff~~~b^{25}=\dfrac{3^{20}}{3^{120}}\\\\\iff~~~b^{25}=3^{20-120}\\\\\iff~~~b^{25}=3^{-100}\\\\\iff~~~b^{25}=\dfrac{1}{3^{100}}\\\\\iff~~~b=\sqrt[25]{\dfrac{1}{3^{100}}}\\\\\iff~~~b=\dfrac{\sqrt[25]{1}}{\sqrt[25]{3^{100}}}\\\\\iff~~~b=\dfrac{1}{\sqrt[25:25]{3^{100:25}}}\\\\\iff~~\boldsymbol{\boxed{b=\dfrac{1}{3^4}}}\end{array}$

Agora para encontrar ‘‘a’’, vamos substituir ‘‘b’’ na eq. em que a isolamos:

$\\\large\begin{array}{l}\implies~~~a=b^4\cdot3^{20}\\\\\iff~~~a=\bigg(\dfrac{1}{3^4}\bigg)\!^4\cdot3^{20}\\\\\iff~~~a=(3^{-4})^4\cdot3^{20}\\\\\iff~~~a=3^{-16}\cdot3^{20}\\\\\iff~~~a=3^{-16\:+\:20}\\\\\iff~~\boldsymbol{\boxed{a=3^{4}}}\end{array}$

Dessa forma, se a = 3⁴ e b = 1/3⁴, então o produto P = a · b é:

$\\\Large\begin{array}{l}P=a\cdot b\\\\P=3^4\cdot\dfrac{1}{3^4}\\\\P=\diagdown\!\!\!\!3^4\cdot\dfrac{1}{\diagdown\!\!\!\!3^4}\\\\\!\!\boldsymbol{\boxed{\boxed{P=1}}}\end{array}$

⠀

$\!\!\!\!\Large\begin{array}{l}\beta\gamma~N\alpha sg\theta v\alpha sk\theta v\\\Huge\text{\sf ---------------------------------------------}\end{array}$

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