Why is 2^(log base 2 of x) equal to x?

[ixbo100]

I'm getting a refresher on logs for my CS class. We're working with log base 2.

I'm having trouble figuring out why 2 ^ (log base 2 of x) = x

Appreciate your help, thank you!

 

[tkhunny]

Couple ways to go about it.

Write it in logarithmic form using this equivalence $\displaystyle log_{b}(a) = c \iff b^{c} = a$

$\displaystyle 2^{log_{2}(x)} = x \iff log_{2}(x) = log_{2}(x)$

Introduce a logarithm using this ides, $\displaystyle If\;a^{b} = c\;then\;log\left(a^{b}\right) = log(c)\;$

$\displaystyle If \;2^{log_{2}(x)} = x \; then \; log_{2}\left(2^{log_{2}(x)}\right) = log_{2}(x)$

There may be other useful demonstrations.

 

[mmm4444bot]

Why is 2^(log base 2 of x) equal to x?

Because logs are exponents.

log₂(x) is the exponent to which 2 must be raised in order for the power to equal x. :cool:

 

[JeffM]

This is very much in the spirit of the other two answers.

$\displaystyle \text {Let } z = 2^{log_2(x)} \implies log_2(z) = log_2 \left ( 2^{log_2(x)} \right ) \implies$

$\displaystyle log_2(z) = log_2(x) * log_2(2) = log_2(x) * 1 = log_2(x) \implies$

$\displaystyle z = x \implies 2^{log_2(x)} = x.$

 

[HallsofIvy]

The real question is what definitions of "exponentials" and "logarithms" are you using?