<s>[INST] Solve for x in the equation: $3^x + 9^x = 27^x$ Solve for x in the equation: $3^x + 9^x = 27^x$ 
Via calculator, the x is = 0.438 which is correct as it is in the choices. However I want to know how to solve it manually, hence the question.
I'm stuck at : $3^x + 3^{2x} = 3^{3x}$
I don't know what to do next. Any help would be appreciated.
 [/INST] $$3^x + 9^x = 27^x$$
$$3^x+3^{2x} = 3^{3x}$$
let $y= 3^x$
$$y+y^2=y^3$$
$$y^3-y^2-y = 0 $$
$$y(y^2-y-1)=0$$
so either $y=0$ or  $(y^2-y-1)=0$ 
since $3^x$  can never be $0$;
$(y^2-y-1)=0$
$\implies y = \dfrac{1\pm\sqrt{5}}{2}$
$\implies 3^x = \dfrac{1\pm\sqrt5}{2}$
$\implies x =\log_3\left(\dfrac{1\pm\sqrt5}{2}\right)$
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