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Recall what $\log_2$ means: $2^P =Q$, $\log_2Q=P$

What is $2^{\log N}$? There is a relationship between $2$ and $\log$, so we should be able to simplify this.

Let $P = 2^{\log N}$. By the definition of $\log_2$ we can write this as $\log_2P = \log_2N$.This means that $P = N$.

Let $P =2^{ \log N}$

$\log_2P = \log_2N$

$P=N$

$2^{\log N}=N$

I'm confused on the line

By the definition of $\log_2$ we can write this as $\log_2P = \log_2N$

If I multiplied $P = 2^{\log N}$ by $\log_2$, then it'd be $\log_2P =\log_2( 2^{\log N})$

$$\log_2P =\log N \cdot \log_2( 2)$$

$$\log_2P =\log N \cdot 1$$

I'm confused on how we can assume that the base of $\log N$ is two?

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    $\begingroup$ This statement only holds if the base of $\log N$ is indeed two, what is implied by the context. The $\log$ symbol can represent base 10, base $e$ or base 2, depending on the context. Also, see that you don't "multiply" by $\log_2$, but you 'apply' this function to both sides of your expression. $\endgroup$
    – rafa11111
    Jun 3, 2018 at 2:59
  • $\begingroup$ You take the $\log_2$ of both sides. The LHS is self explanatory, on the RHS you end up with the exponent of $2$. $\endgroup$
    – Badr B
    Jun 3, 2018 at 3:00

2 Answers 2

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Recall the logarithm properties: $$x=\log_b(b^x)$$ and also $$x=b^{\log_bx}$$

In this case, $2^{\log_2N}$ just equals $N$, because the exponential and logarithmic function cancel out (they're inverses of each other). If $P=2^{\log_2N}$, and we already established that $2^{\log_2N}$ is just $N$, then $P = N$

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$$P =2^{log N}\implies \log_2P=\log N $$

The base for $\log N$ is not specified at this point.

If you assume that $log N$ is base $2$, then you get $P=N$, otherwise you get a different answer.

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