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Quick way to figure out exponents?

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by kishore » Sat Jun 21, 2008 6:54 pm
Ans) 15

2^x - 2^x-2 = 3 * (2 ^ 13)

2^x - 2^x/2 ^2 = 3 * (2 ^ 13)

take 2 ^x as common factor

2^x(1-1/4) = 3 * (2 ^ 13)

2^x * 3/4 = 3 * (2 ^ 13)

3 * 2^x/4 = 3 * (2 ^ 13)

3 * 2^x/2^2 = = 3 * (2 ^ 13)

3 * 2^x-2 = 3 * (2 ^ 13)

x-2 = 13

x = 15
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by rosh26 » Sat Jun 21, 2008 8:31 pm
Can you please explain step #2:

2^x - 2^x/2 ^2 = 3 * (2 ^ 13)

Why is it 2^x/2 ^2 ????

Thanks in advance.
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by shilotilo » Sat Jun 21, 2008 11:27 pm
Hi Rosh,
I did it (after trial and error) in a shorter way:

You have to keep in mind that when two equal bases with different or same powers get multiplied, the powers actually get added.

Hence, 2^x - 2^x-2 can also be re-written as 2^x-2 (2^2 -1)

(In case this step isn't clear, just think, what will happen if powers like x-2 and 2 are added? The 2s get cancelled and you are left with x)

so, what the question actually translates into is 2^x-2 (2^2 -1) =3 (2^13)
or 2^x-2 (4-1) =3 (2^13)
or 2^x-2 (3) =3 (2^13)
Ignore the bases (as they are the same) and solve for x in the powers, and you have the answer, 15.

Shilo
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