Let K be equal to the following expression. What is 2 0 1 4 K ? K = n = 2 ∑ 2 0 1 4 lo g n 2 0 1 4 1
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thanks for an elegant solution "anish puthuraya"
Exactly the solution I was looking for!
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Are there any other methods to solve the problem? I would love to check them out.
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Yours is the one I had in mind. I don't think there is another way to do it. The trick is to justify that lo g a b 1 = lo g b a for a , b = 1 . For completion's sake, can you figure out how to do this?
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@Trevor B. – Well, it is pretty straightforward.
Let lo g a b = k (say)
Then,
b
=
a
k
⇒ b k 1 = a
Taking lo g to the base b on both sides,
⇒ k 1 = lo g b a
Re-substituting the value of k , we get,
lo g a b 1 = lo g b a
Hence, Proved.
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@Anish Puthuraya – Very good! Personally, I would say this.
lo g a b 1 × lo g b a lo g b a = lo g a b × lo g b a lo g b a = 1 lo g b a = lo g b a
You can use change of base to prove that lo g a b × lo g b a = 1 , as I discussed in a note I created a couple weeks ago.
correct
How did the final answer come from the previous step? Please...
what does this ! mean?
I had the same solution but I voted up this not to watse time
Expression for K is to be found . Not of 2014^ K
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Note that,
lo g n 2 0 1 4 1 = lo g 2 0 1 4 n
Thus,
K = n = 2 ∑ 2 0 1 4 lo g 2 0 1 4 n
K = lo g 2 0 1 4 2 + lo g 2 0 1 4 3 + … + lo g 2 0 1 4 2 0 1 4
K = lo g 2 0 1 4 ( 2 × 3 × 4 × … × 2 0 1 4 )
K = lo g 2 0 1 4 2 0 1 4 !
Therefore,
2 0 1 4 K = 2 0 1 4 lo g 2 0 1 4 2 0 1 4 ! = 2 0 1 4 !