Answer
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Hint: Divide 256 by 2. Then divide the quotient that you get by 2. Then again divide the quotient that you get by 2. Keep on doing this until finally, the quotient is 2 itself. Then write all those 2’s together in the power of 2. That will be your answer.
Complete step-by-step answer:
Let us say we want to write $ b $ as a power of $ a $ . Then we should first divide $ b $ by $ a $ and get a quotient $ c $ .
Then we can write,
$ b = ac $
Then we should divide $ c $ by $ a $ and get a quotient $ d $ .
Then we can write,
$ c = ad $
Hence, we can write $ b $ as
$ b = ac = a(ad) = {a^2}d $
Then we should divide $ d $ by $ a $ and get a quotient $ e $ .
Then we can write,
$ d = ae $
Hence, we can again write $ b $ as
$ b = {a^2}d = {a^2}(ae) = {a^3}e $
Then we should divide $ e $ by $ a $ and so on.
This process should go on until finally, the quotient we get is $ a $ itself.
If this process is done $ n $ times till you get $ a $ as a quotient. Then your answer will be
$ b = {a^n} $
In this question,
$ b = 256 $ and $ a = 2 $
So, by using the logic explained above, we can write
$ 256 = 2 \times 128 $
$ \Rightarrow 256 = 2 \times 2 \times 64 $ . . . (1)
$ \Rightarrow 256 = {2^2} \times 2 \times 32 $
$ \Rightarrow 256 = {2^3} \times 2 \times 16 $
$ \Rightarrow 256 = {2^4} \times 2 \times 8 $
$ \Rightarrow 256 = {2^5} \times 2 \times 4 $
$ \Rightarrow 256 = {2^6} \times 2 \times 2 $
$ \Rightarrow 256 = {2^8} $
So, the correct answer is “ $ {2^8} $ ”.
Note: It is not always necessary to keep on dividing by the same number. This is the concept that we use. But if we know the powers of some numbers that we could use to simplify the question in less steps, then we can use that too. For example, in this question, after equation (1), we know that $ 64 = {4^3} $ .
So we can simply further as
$ 256 = 2 \times 2 \times {4^3} $
$ \Rightarrow 256 = {2^2} \times {({2^2})^3} $
$ \Rightarrow 256 = {2^2} \times {2^6} = {2^8} $
Or you if you know that $ 256 = {16^2} $ . Then you can also simplify it as
$ 256 = {16^2} = {({4^2})^2} = {({2^2})^4} = {2^8} $
So, we can reduce our steps if we know some shortcuts and have good knowledge of exponents. Otherwise. The logic explained in the solution will work for any number.
Complete step-by-step answer:
Let us say we want to write $ b $ as a power of $ a $ . Then we should first divide $ b $ by $ a $ and get a quotient $ c $ .
Then we can write,
$ b = ac $
Then we should divide $ c $ by $ a $ and get a quotient $ d $ .
Then we can write,
$ c = ad $
Hence, we can write $ b $ as
$ b = ac = a(ad) = {a^2}d $
Then we should divide $ d $ by $ a $ and get a quotient $ e $ .
Then we can write,
$ d = ae $
Hence, we can again write $ b $ as
$ b = {a^2}d = {a^2}(ae) = {a^3}e $
Then we should divide $ e $ by $ a $ and so on.
This process should go on until finally, the quotient we get is $ a $ itself.
If this process is done $ n $ times till you get $ a $ as a quotient. Then your answer will be
$ b = {a^n} $
In this question,
$ b = 256 $ and $ a = 2 $
So, by using the logic explained above, we can write
$ 256 = 2 \times 128 $
$ \Rightarrow 256 = 2 \times 2 \times 64 $ . . . (1)
$ \Rightarrow 256 = {2^2} \times 2 \times 32 $
$ \Rightarrow 256 = {2^3} \times 2 \times 16 $
$ \Rightarrow 256 = {2^4} \times 2 \times 8 $
$ \Rightarrow 256 = {2^5} \times 2 \times 4 $
$ \Rightarrow 256 = {2^6} \times 2 \times 2 $
$ \Rightarrow 256 = {2^8} $
So, the correct answer is “ $ {2^8} $ ”.
Note: It is not always necessary to keep on dividing by the same number. This is the concept that we use. But if we know the powers of some numbers that we could use to simplify the question in less steps, then we can use that too. For example, in this question, after equation (1), we know that $ 64 = {4^3} $ .
So we can simply further as
$ 256 = 2 \times 2 \times {4^3} $
$ \Rightarrow 256 = {2^2} \times {({2^2})^3} $
$ \Rightarrow 256 = {2^2} \times {2^6} = {2^8} $
Or you if you know that $ 256 = {16^2} $ . Then you can also simplify it as
$ 256 = {16^2} = {({4^2})^2} = {({2^2})^4} = {2^8} $
So, we can reduce our steps if we know some shortcuts and have good knowledge of exponents. Otherwise. The logic explained in the solution will work for any number.
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