How do you solve $x^{2} 3^{x} - 8 (3^{x}) = 0$ $x^2 3^x -8(3^x)= 0$ ?

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How do you solve $x^{2} 3^{x} - 8 (3^{x}) = 0$ $x^2 3^x -8(3^x)= 0$ ?

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Answer 1 · 2018-04-21T09:02:20

$x = \pm 2 \sqrt{2}$ $x=+-2sqrt2$

Explanation:

$factor out 3^{x}$ $"factor out "3^x$

$\Rightarrow 3^{x} (x^{2} - 8) = 0$ $rArr3^x(x^2-8)=0$

$equate each factor to zero and solve for x$ $"equate each factor to zero and solve for x"$

$note that 3^{x} \neq 0$ $"note that "3^x!=0$

$\Rightarrow x^{2} - 8 = 0 \Rightarrow x^{2} = 8 \Rightarrow x = \pm 2 \sqrt{2} is the solution$ $rArrx^2-8=0rArrx^2=8rArrx=+-2sqrt2" is the solution"$

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How do you solve $x^{2} 3^{x} - 8 (3^{x}) = 0$ $x^2 3^x -8(3^x)= 0$ ?

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How do you solve $x^{2} 3^{x} - 8 (3^{x}) = 0$ $x^2 3^x -8(3^x)= 0$ ?