Algebra of exponential

Question

Solve for $x$ in exact value:

$\\3^{2x}-3^{x+2}+8=0$

I have tried substituting $3^x$ $=a$ but I didn't get anywhere.

$\\a^2-a^{1+\frac{2}{x}}+8=0$

Answer 1

$$3^{2x}−3^2*3^x+8=0$$ Substituting $3^x =a$: $$a^2-3^2*a+8=0$$ $$a^2-9a+8=0$$ $$(a-1)(a-8)=0$$ $$a_1=1, a_2=8$$ Final answer: $$x_1=0, x_2=log_38$$

Answer 2

Notice that $\log_x(y)=z\Longleftrightarrow x^z=y$

$$8+3^{2x}-3^{x+2}=0\Longleftrightarrow$$

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Substitute $y=3^x$:

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$$y^2-9y+8=0\Longleftrightarrow$$ $$(y-9)(y-1)=0$$

So we got:

• $$y-9=0\Longleftrightarrow$$ $$y=9\Longleftrightarrow$$ $$3^x=9\Longleftrightarrow$$ $$x=\frac{\ln(8)}{\ln(3)}\Longleftrightarrow$$ $$x=\log_3(8)$$
• $$y-1=0\Longleftrightarrow$$ $$y=1\Longleftrightarrow$$ $$3^x=1\Longleftrightarrow$$ $$x=0$$