Question

Equação logarítmica 5 exercícios:
(Eu já possuo as respostas, mas preciso da resolução passo a passo, obg)

1. Log 3 x + Log 9 x = 3

2. Log 2 x + Log 16 x = 3

3. Log 2 x + Log 4 x = 9

4. Log 3 (x – 2) + Log 9 (x – 2 ) = 3

5. Log 4 (x + 2) + Log 2 (x + 2 ) = 3

Answer 1

Note que em cada um dos 5 exercícios, temos dois logaritmos de mesmo logaritmando e bases diferentes. Dessa forma, para podermos resolver as equações podemos mudar a base de um dos dois logaritmos para coincidir com a base do outro.

Para isso vamos utilizar a propriedade da troca de base:

$\boxed{\log_{_b}a~=~\dfrac{\log_{_c}a}{\log_{_c}b}}$

Lembre-se que ao final será preciso verificar se a solução encontrada (valor de "x") atende às condições de existência dos logaritmos (C.E's).

$Seja~~ \log_{_b}a,~~C.E.:~~\left\{\begin{array}{ccc}a&>&0\\b&>&0\\b&\ne&1\end{array}\right$

1.

$\log_{_3}x~+~\log_{_9}x~=~3\\\\\\Mudando~a~base~do~segundo~\log~de~9~para~3:\\\\\\\log_{_3}x~+~\dfrac{\log_{_3}x}{\log_{_3}9}~=~3\\\\\\Fatorando~o~logaritmando~9\\\\\\\log_{_3}x~+~\dfrac{\log_{_3}x}{\log_{_3}3^2}~=~3\\\\\\Aplicando~a~propriedade~do~logaritmo~da~potencia\\\\\\\log_{_3}x~+~\dfrac{\log_{_3}x}{2\cdot\log_{_3}3}~=~3\\\\\\\log_{_3}x~+~\dfrac{\log_{_3}x}{2\cdot1}~=~3\\\\\\\log_{_3}x~+~\dfrac{1}{2}\cdot\log_{_3}x~=~3\\\\\\\log_{_3}x\cdot \left(1~+~\dfrac{1}{2}\right)~=~3$

$\log_{_3}x\cdot \left(\dfrac{3}{2}\right)~=~3\\\\\\\log_{_3}x~=~\dfrac{3}{\frac{3}{2}}\\\\\\\log_{_3}x~=~2\\\\\\x~=~3^2\\\\\\\boxed{x~=~9}~~~\Rightarrow~Atende~\grave{a}s~C.E.~~\checkmark$

2.

$\log_{_2}x~+~\log_{_{16}}x~=~3\\\\\\Mudando~a~base~do~segundo~\log~de~16~para~2:\\\\\\\log_{_2}x~+~\dfrac{\log_{_2}x}{\log_{_2}16}~=~3\\\\\\Fatorando~o~logaritmando~16\\\\\\\log_{_2}x~+~\dfrac{\log_{_2}x}{\log_{_2}2^4}~=~3\\\\\\Aplicando~a~propriedade~do~logaritmo~da~potencia\\\\\\\log_{_2}x~+~\dfrac{\log_{_2}x}{4\cdot\log_{_2}2}~=~3\\\\\\\log_{_2}x~+~\dfrac{\log_{_2}x}{4\cdot1}~=~3\\\\\\\log_{_2}x~+~\dfrac{1}{4}\cdot\log_{_2}x~=~3\\\\\\\log_{_2}x\cdot \left(1~+~\dfrac{1}{4}\right)~=~3$

$\log_{_2}x\cdot \left(\dfrac{5}{4}\right)~=~3\\\\\\\log_{_2}x~=~\dfrac{3}{\frac{5}{4}}\\\\\\\log_{_2}x~=~\dfrac{12}{5}\\\\\\\boxed{x~=~2^{\frac{12}{5}}}~~ou~~\boxed{x~=~\sqrt[5]{4096}}~~~\Rightarrow~Atende~\grave{a}s~C.E.~~\checkmark$

3.

$\log_{_2}x~+~\log_{_4}x~=~9\\\\\\Mudando~a~base~do~segundo~\log~de~4~para~2:\\\\\\\log_{_2}x~+~\dfrac{\log_{_2}x}{\log_{_2}4}~=~9\\\\\\Fatorando~o~logaritmando~4\\\\\\\log_{_2}x~+~\dfrac{\log_{_2}x}{\log_{_2}2^2}~=~9\\\\\\Aplicando~a~propriedade~do~logaritmo~da~potencia\\\\\\\log_{_2}x~+~\dfrac{\log_{_2}x}{2\cdot\log_{_2}2}~=~9\\\\\\\log_{_2}x~+~\dfrac{\log_{_2}x}{2\cdot1}~=~9\\\\\\\log_{_2}x~+~\dfrac{1}{2}\cdot\log_{_2}x~=~9\\\\\\\log_{_2}x\cdot \left(1~+~\dfrac{1}{2}\right)~=~9$

$\log_{_2}x\cdot \left(\dfrac{3}{2}\right)~=~9\\\\\\\log_{_2}x~=~\dfrac{9}{\frac{3}{2}}\\\\\\\log_{_2}x~=~6\\\\\\x~=~2^6\\\\\\\boxed{x~=~64}~~~\Rightarrow~Atende~\grave{a}s~C.E.~~\checkmark$

4.

$\log_{_3}(x-2)~+~\log_{_9}(x-2)~=~3\\\\\\Mudando~a~base~do~segundo~\log~de~9~para~3:\\\\\\\log_{_3}(x-2)~+~\dfrac{\log_{_3}(x-2)}{\log_{_3}9}~=~3\\\\\\Fatorando~o~logaritmando~9\\\\\\\log_{_3}(x-2)~+~\dfrac{\log_{_3}(x-2)}{\log_{_3}3^2}~=~3\\\\\\Aplicando~a~propriedade~do~logaritmo~da~potencia\\\\\\\log_{_3}(x-2)~+~\dfrac{\log_{_3}(x-2)}{2\cdot\log_{_3}3}~=~3\\\\\\\log_{_3}(x-2)~+~\dfrac{\log_{_3}(x-2)}{2\cdot1}~=~3\\\\\\\log_{_3}(x-2)~+~\dfrac{1}{2}\cdot\log_{_3}(x-2)~=~3$

$\log_{_3}(x-2)~+~\dfrac{1}{2}\cdot\log_{_3}(x-2)~=~3\\\\\\\log_{_3}(x-2)~+~\left(1+\dfrac{1}{2}\right)~=~3\\\\\\\log_{_3}(x-2)\cdot \left(\dfrac{3}{2}\right)~=~3\\\\\\\log_{_3}(x-2)~=~\dfrac{3}{\frac{3}{2}}\\\\\\\log_{_3}(x-2)~=~2\\\\\\x-2~=~3^2\\\\\\x~=~9+2\\\\\\\boxed{x~=~11}~~~\Rightarrow~Atende~\grave{a}s~C.E.~~\checkmark$

5.

$\log_{_4}(x+2)~+~\log_{_2}(x+2)~=~3\\\\\\Mudando~a~base~do~primeiro~\log~de~4~para~2:\\\\\\\dfrac{\log_{_2}(x+2)}{\log_{_2}4}~+~\log_{_2}(x+2)~=~3\\\\\\Fatorando~o~logaritmando~4\\\\\\\dfrac{\log_{_2}(x+2)}{\log_{_2}2~2}~+~\log_{_2}(x+2)~=~3\\\\\\Aplicando~a~propriedade~do~logaritmo~da~potencia\\\\\\\dfrac{\log_{_2}(x+2)}{2\cdot\log_{_2}2}~+~\log_{_2}(x+2)~=~3\\\\\\\dfrac{\log_{_2}(x+2)}{2\cdot1}~+~\log_{_2}(x+2)~=~3\\\\\\\dfrac{1}{2}\cdot \log_{_2}(x+2)~+~\log_{_2}(x+2)~=~3$

$\log_{_2}(x+2)~+~\left(\dfrac{1}{2}+1\right)~=~3\\\\\\\log_{_2}(x+2)~+~\left(\dfrac{3}{2}\right)~=~3\\\\\\\log_{_2}(x+2)~=~\dfrac{3}{\frac{3}{2}}\\\\\\\log_{_2}(x+2)~=~2\\\\\\x+2~=~2^2\\\\\\x~=~4-2\\\\\\\boxed{x~=~2}~~\Rightarrow~Atende~\grave{a}s~C.E.~\checkmark\\\\\\\Huge{\begin{array}{c}\Delta \tt{\!\!\!\!\!\!\,\,o}\!\!\!\!\!\!\!\!\:\,\perp\end{array}}Qualquer~d\acute{u}vida,~deixe~ um~coment\acute{a}rio$