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?
Best Answer
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 $P = N$