Why is 2^(log base 2 of x) equal to x?
- Thread starter ixbo100
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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.
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.
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mmm4444bot
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Because logs are exponents.
log2(x) is the exponent to which 2 must be raised in order for the power to equal x. :cool:
This is very much in the spirit of the other two answers.
\(\displaystyle \text {Let } z = 2^{log_2(x)} \implies log_2(z) = log_2 \left ( 2^{log_2(x)} \right ) \implies\)
\(\displaystyle log_2(z) = log_2(x) * log_2(2) = log_2(x) * 1 = log_2(x) \implies\)
\(\displaystyle z = x \implies 2^{log_2(x)} = x.\)
\(\displaystyle \text {Let } z = 2^{log_2(x)} \implies log_2(z) = log_2 \left ( 2^{log_2(x)} \right ) \implies\)
\(\displaystyle log_2(z) = log_2(x) * log_2(2) = log_2(x) * 1 = log_2(x) \implies\)
\(\displaystyle z = x \implies 2^{log_2(x)} = x.\)
HallsofIvy
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The real question is what definitions of "exponentials" and "logarithms" are you using?