# Solving an Exponential Equation

Answer is given, and it equals to 1.

$$2\cdot 16^{x}-2^{4x}-4^{2x-2}=15$$

$2^{4x-4}=-6,5$ <- This is where I reached, which is clearly wrong

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You given answer clearly doesn't solve the equation you reached (2^0 = 1), nor does it solve the initial equation. $2\times 16^1 - 2^1-4^{0} = 29$. Also, there is no question here. – Mikael Öhman Aug 27 '11 at 14:20
@Mikael: I pointed this out already, and BeatShot has corrected the statement. I agree with you that Beatshot should state the question clearly, but it is certainly along the lines of "how to proceed?" – Zev Chonoles Aug 27 '11 at 14:22
@Mikael Öhman, I know the point I reached with the equation is wrong. – BeatShot Aug 27 '11 at 14:24
@BeatShot: The edit had not shown up when I posted my comment. – Mikael Öhman Aug 27 '11 at 14:31

Note that $$2\cdot 16^x=2\cdot (2^4)^x=2\cdot2^{4x}=2^{4x+1}$$ and $$4^{2x-2}=(2^2)^{2x-2}=2^{4x-4}.$$ So you want to solve $$2\cdot16^x-2^{4x}-4^{2x-2}=15$$ $$2^{4x+1}-2^{4x}-2^{4x-4}=15$$ See any good ways of factoring something out on the left side of the equation? I think you might have reached this stage but made a minor calculation error. If you want to check your work, here is a spoiler (if you put your cursor over the box, the next step will appear):
$$2^{4x-4}(2^5-2^4-1)=15$$
Instead of $2^{4x+1}$, I got $2^{4x}$ and 15/2 to the right side. – BeatShot Aug 27 '11 at 14:32
Ah, in these kinds of problems it's usually best to avoid trying to cancel things from both sides prematurely - keep all the 2's in the powers until you can find some other way of solving. Especially in this case, because 2 doesn't go into 15, writing $\frac{15}{2}$ doesn't simplify matters. – Zev Chonoles Aug 27 '11 at 14:39