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Re: If 2^x - 2^(x-2) = 3*2^(13), what is x? [#permalink]
28 Aug 2013, 13:27

I did it a bit differently, but I arrived at the right answer. Here is the way I did it: \(2^x - 2^{x-2} = 3* 2^{13}\) Factor out a \(2^x\) which gives: \(2^x(1 - \frac{1}{4}) = 3* 2^{13}\) Clean up: \(2^x(\frac{3}{4}) = 3* 2^{13}\) At this point I realized that the 4 in the denominator could be factored out so that's what I did: \(2^{x-2}(3) = 3* 2^{13}\) From here you have \(2^{x-2} = 2^{13}\) so \(x = 15\)

Can you please explain the red highlighted part? I read other explanations but it wasn't clear. I am unable to understand how did you factor out 2^(x-2)?

Can you please explain the red highlighted part? I read other explanations but it wasn't clear. I am unable to understand how did you factor out 2^(x-2)?

Sure! First notice that \(2^{x-2} * 2^{2} = 2^{x}\) So, we know that \(2^{x-2}\) is a factor of \(2^{x}\). I am using the product rule for exponents: \(x^{a}*x^{b}=x^{a+b}\) It helps to think of this rule in reverse (going from right -> left). What I mean by that is we can also write it as \(x^{a+b}=x^{a}*x^{b}\) When I factor out the \(2^{x-2}\) I am really separating \(2^{x}\) into \(2^{x-2} * 2^{2}\). So, \(2^x - 2^{x-2} = 3*2^{13}\) which becomes \(2^{x-2}(2^2-1)= 3*2^{13}\) after we factor out the \(2^{x}\).

Re: If 2^x - 2^(x-2) = 3*2^(13), what is x? [#permalink]
09 Sep 2013, 08:30

Quote:

Clean up: 2^x(\frac{3}{4}) = 3* 2^{13} At this point I realized that the 4 in the denominator could be factored out so that's what I did: 2^{x-2}(3) = 3* 2^{13} From here you have 2^{x-2} = 2^{13} so x = 15

It seems that in the bolded step above you could have multiplied both sides by 4/3 thus canceling the 3 from the other side out. You would then be left with 4 or (2*2) or 2^2 in addition to the 2^13 leaving you with the 2^15. Just thought that might be a little quicker. _________________

Now it's a pretty simple point, but it got silly old me baffled, so i searched up the net and came up with the following explanation. \(2^{x-2}(2^2-1)= 3*2^{13}\)

If from \(N^x - N^{x-a}, N^{x-a}\) is factored out we will have: \(N^{x-a} (N^b - 1)\) Where \(b = x - a\) Example 1 \(5^8 - 5^5 = 5^5 (5^3 - 1) = 5^5 (125 - 1) = 5^5*124\) Example 2 \(2^x - 2^{x-2} = 2^x (2^2 - 1) = 2^x (4 - 1) = 2^x . 3\)

Now in Example \(2\) we don't know the value of the \(x\) but we know the difference between\(x\) and\(x-2\) is \(2\) therefore,\(b = 2\).

Now it's a pretty simple point, but it got silly old me baffled, so i searched up the net and came up with the following explanation. \(2^{x-2}(2^2-1)= 3*2^{13}\)

If from \(N^x - N^{x-a}, N^{x-a}\) is factored out we will have: \(N^{x-a} (N^b - 1)\) Where \(b = x - a\) Example 1 \(5^8 - 5^5 = 5^5 (5^3 - 1) = 5^5 (125 - 1) = 5^5*124\) [color=#ff0000]Example 2 [b]\(2^x - 2^{x-2} = 2^x (2^2 - 1) = 2^x (4 - 1) = 2^x . 3\)[/color][/b]

Now in Example \(2\) we don't know the value of the \(x\) but we know the difference between\(x\) and\(x-2\) is \(2\) therefore,\(b = 2\).

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