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I solved this , but I am not sure if I did in the right way.

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

$$2^{x + 2} - 2^{2x + 2} = 8$$

$$\log_22^{x + 2} - \log_22^{2x + 2} = \log_28$$

$$x + 2- 2x - 2 = 3$$

solving for $x$:

$$x = -2$$

any feedback would be appreciated.

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That answer cannot work, you can substitute and check. You are assuming $\log (a - b) = \log a - \log b$ and then that $\log x - \log y = \log (x-y)$, but both steps are not valid. –  Macavity Dec 7 '13 at 1:58
    
Another variant of the freshman's dream. –  Michael Hoppe Dec 7 '13 at 2:01

3 Answers 3

up vote 1 down vote accepted

You can do it without including logarithms . Just take $t = 2^x$. It will become a quadratic in $t$ solve for $t$ . If $t$ has any negative value , neglect it beacuse $t> 0$ .

$t^2-2t+4 = 0 $ which has complex roots and hence there is no solution for $x$ in the real domain .

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well, but I dont seem to get there either by substitution, could show me how you'd solve it ? –  victor Dec 7 '13 at 2:07

You can easily check the solution you get. Checking $x=-2$, you get $$ 2^{2x+1}-2^{x+2}+8=2^{-3}-2^{0}+8=\dfrac{1}{8}-1+8\ne0, $$which means you have an error. Specifically, you transition from the first line to the second line is incorrect.

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let $2^{x+2}=t$ then $2^{2x+1}=\frac{1}{8}t^2$

substituting in the equation ,you get $$\frac{1}{8}t^2-t+8=0$$ which is a quadratic equation and can be easily solved.

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