Solve for x in the equation: $3^x + 9^x = 27^x$

Question

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.

Answer 1

When there is something in common across an expression, such as $3^x$ is here, it is helpful to think of that quantity as a "sub-expression". Starting with

$$3^x+3^{2x}=3^{3x}\\ 3^x+(3^x)^2=(3^x)^3$$

the next step would be to create that sub-expression as $y=3^x$, thus giving $y+y^2=y^3$. $y=0$ is not viable as $3^x=0$ does not have a solution for real $x$, so then we must solve $1+y=y^2$ (getting $y=\frac {1\pm\sqrt 5}2$, only one value of which is viable...) and substitute in reverse to get a value for $x$, as $x=\log_3y$. $y=\frac {1-\sqrt 5}2$ is not viable since the value is negative and there are again no real solutions for $3^x=y$, so the answer is $x=\log\frac{1+\sqrt 5}2=\log(1+\sqrt 5)-\log 2$.

Answer 2

Next step is to rewrite $$ (3^x)+(3^x)^2-(3^x)^3=0\\ 3^x(1+(3^x)-(3^x)^2)=0\\ y(1+y-y^2)=0 $$ where $y=3^x$. Find $y$, then find $x$.

Answer 3

$$3^x + 9^x = 27^x$$

$$3^x+3^{2x} = 3^{3x}$$

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 $0$;

$(y^2-y-1)=0$

$\implies y = \dfrac{1\pm\sqrt{5}}{2}$

$\implies 3^x = \dfrac{1\pm\sqrt5}{2}$

$\implies x =\log_3\left(\dfrac{1\pm\sqrt5}{2}\right)$

Answer 4

Substitute 3^x=t
then we can get the equation as t+t^2=t^3
AS 3^x not equal 0 then the eqn is t^2-t-1=0
we can easily solve the eqn and get the answer t=1.618
So 3^x=1.618 therefore x=log 3 1.618
x=0.438