Question

Dê o valor real de x que satisfaça a seguinte equação : 4^x + 6^x = 9^x

Answer 1

$\mathsf{{4}^{x}+{6}^{x}={9}^{x}}\\\mathsf{\dfrac{{4}^{x}}{{9}^{x}}+\dfrac{{6}^{x}}{{9}^{x}}=1}$

$\mathsf{{[{(\dfrac{2}{3})}^{2}] }^{x}+{(\dfrac{2}{3})}^{x}=1}$

$fazendo\,{(\dfrac{2}{3})}^{x}=y\,temos$

$\mathsf{{y}^{2}+y=1}\\\mathsf{{y}^{2}+y-1=0}$

$\mathsf{\Delta=1+4=5}\\\mathsf{y=\dfrac{-1+\sqrt{5}}{2}}$

$\mathsf{{(\dfrac{2}{3})}^{x}=\dfrac{-1+\sqrt{5}}{2}}$

$\large\mathsf{x=log_{\frac{2}{3}}(\dfrac{-1+\sqrt{5}}{2})}$