​What are the values of $$x$$ in the following equation?

$$3^{2x+1} - 3^x = 3^{x+3} - 3^2$$​​

Correct option is C

Given:

$$3^{2x+1} - 3^x = 3^{x+3} - 3^2$$

Formula Used:

$$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$​

Solution:

​$$3^{2x+1} - 3^x = 3^{x+3} - 3^2$$

$$3 \cdot 3^{2x} - 3^x - 27 \cdot 3^x + 9 = 0$$

$$3 \cdot 3^{2x} - 28 \cdot 3^x + 9 = 0$$

Put; $$y = 3^x$$

$$3^{2x} = (3^x)^2 = y^2$$

Then,

$$3y^2 - 28y + 9 = 0$$

Here, a = 3, b = -28 , c = 9

$$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

$$y = \frac{-(-28) \pm \sqrt{(-28)^2 - 4(3)(9)}}{2(3)}$$

$$y = \frac{28 \pm \sqrt{784 - 108}}{6}$$

$$y = \frac{28 \pm \sqrt{676}}{6}$$

$$y = \frac{28 \pm 26}{6}$$

$$y = \frac{28 + 26}{6} = \frac{54}{6} = 9$$ ;  $$y = \frac{28 - 26}{6} = \frac{2}{6} = \frac{1}{3}$$

Put; $$y = 3^x$$​​

​$$3^x = 9 \implies 3^x = 3^2 \implies x = 2$$

$$3^x = \frac{1}{3} \implies 3^x = 3^{-1} \implies x = -1$$

Thus, option (c) 2, -1 is right.

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