Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
1  
@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
add comment

1 Answer

up vote 4 down vote accepted

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):

$$2^{4x-4}(2^5-2^4-1)=15$$

share|improve this answer
    
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
    
Thank you for your help, finally solved it! =) –  BeatShot Aug 27 '11 at 14:45
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.