Question

Calcule o valor de x :
$\Large\sf {4}^{x} + {6}^{x} = {9}^{x}$​

Answer 1

Oie, Td Bom?!

$4 {}^{x} + 6 {}^{x} = 9 {}^{x}$

$4 {}^{x} + 6 {}^{x} - 9 {}^{x} = 0$

$(2{}^{2} ) {}^{x} + (2 \: . \: 3) {}^{x} - (3 {}^{2} ) {}^{x}$

$2 {}^{2x} + (2 \: . \: 3) {}^{x} - 3 {}^{2x} = 0$

$2 {}^{2x} + 2 {}^{x} \: . \: 3 {}^{x} - 3 {}^{2x} = 0$

• Divida tudo por $3{}^{2x}$.

$( \frac{2}{3} ) {}^{2x} + ( \frac{2}{3} ) {}^{x} - 1 = 0$

$(( \frac{2}{3} ) {}^{x} ) {}^{2} + ( \frac{2}{3} ) {}^{x} - 1 = 0$

• Substitua $( \frac{2}{3} ) {}^{x} = t$.

$t {}^{2} + t - 1 = 0$

• Coeficientes:

$a = 1 \: , \: b = 1 \: ,\: c = - 1$

• Fórmula resolutiva:

$x = \frac{ - b± \sqrt{b {}^{2} - 4ac } }{2a}$

$t= \frac{ - 1± \sqrt{1 {}^{2} - 4 \: . \: 1 \: . \: ( - 1)} }{2 \: . \: 1}$

$t= \frac{ - 1± \sqrt{1 + 4} }{2}$

$t= \frac{ - 1± \sqrt{5} }{2}$

• Substitua $t = ( \frac{2}{3} ) {}^{x}$.

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

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

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

• $( \frac{2}{3} ) {}^{x} = \frac{ - 1 - \sqrt{5} }{2}$

$x∉\mathbb{R}$

$S = \left \{ log_{ \frac{2}{3} }( \frac{ - 1 + \sqrt{5} }{2} ) \right \}$

Att. Makaveli1996

Answer 2

EQUAÇÃO

                       LOGARITMA

$\sf \boldsymbol{\sf 4^x+6^x=9^x}\\\\\\\\\sf \boldsymbol{\sf }$

• Divida ambos os lados por 4^x

$\sf \displaystyle \boldsymbol{\sf \frac{4^x}{4^x}+\frac{6^x}{4^x}=\frac{9^x}{4^x}}$

• Simplifique

$\sf \displaystyle \boldsymbol{\sf 1+\left(\frac{3}{2}\right)^x=\left(\frac{9}{4}\right)^x}$

• Agora aplique a propriedade dos expoentes.

$\sf \displaystyle \boldsymbol{\sf 1+\left(\frac{3}{2}\right)^x=\left(\left(\frac{3}{2}\right)^x\right)^2}$

• Reescreva a equação com (3/2)^x=u

$\sf \displaystyle \boldsymbol{\sf 1+u=\left(u\right)^2}$

• Resolver 1+u=(u)^2

$\sf \displaystyle \boldsymbol{\sf u=\frac{1+\sqrt{5}}{2},\:u=\frac{1-\sqrt{5}}{2}}$

• Substitua u=(3/2)^x, e solucione x

$\sf \displaystyle \boldsymbol{\sf \left(\frac{3}{2}\right)^x=u=\frac{1+\sqrt{5}}{2}\Rightarrow x=\frac{\ln \left(\frac{1+\sqrt{5}}{2}\right)}{\ln \left(\frac{3}{2}\right)}}\\\\\\\\ \boldsymbol{\sf \left(\frac{3}{2}\right)^xu=\frac{1-\sqrt{5}}{2}\Rightarrow x\in\mathbb{R}}$

$\sf \therefore$  $\boxed{\sf S=\{\sf x=\frac{\ln \left(\frac{1+\sqrt{5}}{2}\right)}{\ln \left(\frac{3}{2}\right)}\approx 6811,18\cdots \}}$

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