Question

equação logarítmica
resolva a equação logx na base 2 + logx na base 3 + logx na base 4=1

Answer 1

Definição

$\boxed{\boxed{\log_{b}a=c~~\Longleftrightarrow~~b^{c}=a~~\mathsf{para~a,b~\textgreater~0~~e~~b\neq1}}}$

Propriedade 1

$\boxed{\boxed{\log_{b^{n}}a=\dfrac{1}{n}\log_{b}a}}$

Mudança de base ($b\to c$)

$\boxed{\boxed{\log_{b}a=\dfrac{\log_{c}a}{\log_{c}b}}}$
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$\log_{2}x+\log_{3}x+\log_{4}x=1\\\\\log_{2}x+\log_{3}x+log_{(2^{2})}x=1$

Usando a propriedade 1 no terceiro logaritmo:

$\log_{2}x+\log_{3}x+\frac{1}{2}\log_{2}x=1\\\\\big(\log_{2}x+\frac{1}{2}\log_{2}x)+\log_{3}x=1\\\\(1+\frac{1}{2})\log_{2}x+\log_{3}x=1\\\\\frac{3}{2}\log_{2}x+\log_{3}x=1$

Multiplicando todos os membros por 2:

$3\log_{2}x+2\log_{3}x=1$

Usando mudança de base (para base 10) nos dois logaritmos:

$\dfrac{3\log x}{\log2}+\dfrac{2\log x}{\log3}=2\\\\\\\bigg(\dfrac{3}{\log2}+\dfrac{2}{\log3}\bigg)\cdot\log x=2\\\\\\\bigg(\dfrac{3\log3+2\log2}{\log2\cdot\log3}\bigg)\cdot\log x=2\\\\\\\bigg(\dfrac{\log3^{3}+\log2^{2}}{\log2\cdot\log3}\bigg)\log x=2\\\\\\\bigg(\dfrac{\log(27\cdot4)}{\log2\cdot\log3}\bigg)\log x=2\\\\\\\bigg(\dfrac{\log108}{\log2\cdot\log3}\bigg)\log x=2$

Isolando $\log x$:

$\log x=\dfrac{2\cdot\log2\cdot\log3}{\log108}\\\\\\\log x=\dfrac{\log2^{2}\cdot\log3}{\log108}\\\\\\\log x=\dfrac{\log4\cdot\log3}{\log108}$

Aplicando a definição de logaritmos, devemos ter

$x=10^{\log4\cdot\log3/\log108}\\\\x=(10^{\log4})^{\log3/\log108}\\\\x=4^{\log3/\log108}\\\\x=(2^{2})^{\log3/\log108}\\\\x=2^{2\log3/\log108}\\\\x=2^{\log3^{2}/\log10}\\\\x=2^{\log9/\log108}$

Podemos escrever $\log_{108}9=\dfrac{\log9}{\log108}$. Então:

$\boxed{\boxed{x=2^{\log_{108}9}}}$

é a solução da equação