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$8^X + 7 (2^{X+1}) = 7 (4^X) + 8$

$2^{3X} + 7 (2^{X+1}) = 7 (2^{2X}) + 2^3$

$3X = 3$

$X =1$

OR

$X+1 = 2X$

$X=1$

BUT answer $X = 0$, or $1$ or $2$ ????

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  • $\begingroup$ What did you do? Also LaTeX it. $\endgroup$ – UserX Oct 7 '14 at 12:56
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I guess you compare $2^{3X}$ with $2^3$ to get $3X=3$ and compare $2^{X+1}$ with $2^{2X}$ to get $X+1=2X$, but you cannot do this in general.

Let $t=2^X$. Then, noting that $$2^{3X}=(2^X)^3=t^3,\ 7\cdot 2^{X+1}=7\cdot 2\cdot 2^X=14t,\ 2^{2X}=(2^X)^2=t^2,$$we have $$t^3-7t^2+14t-8=0\Rightarrow (t-1)(t-2)(t-4)=0\Rightarrow X=0,1,2.$$

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  • $\begingroup$ how to convert t3−7t2+14t−8 to (t−1)(t−2)(t−4) $\endgroup$ – sekling Oct 7 '14 at 12:59
  • $\begingroup$ @sekling: Do you know factor theorem? $\endgroup$ – mathlove Oct 7 '14 at 13:01
  • $\begingroup$ Great, got it, THANKS $\endgroup$ – sekling Oct 7 '14 at 13:05
  • $\begingroup$ @sekling: You are welcome. By the way, I added a bit. Take a look. $\endgroup$ – mathlove Oct 7 '14 at 13:07

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