Quick way to figure out exponents?
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Had to do this the long way on the test...
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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
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
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
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