Logarithms doubt

[JPJ]

Why can I say that \(\displaystyle 2^(ln x) = x^(ln 2)\) ?


ps.: Sorry about the latex typos, couldn't get it to work...meant it like 2^(ln x) = x^(ln 2).

 

[JPJ]

Nevermind about it...just figured it out: 2^lnx = (e^Ln2)^Lnx = e^(Ln2*Lnx) = (e^Lnx)^Ln2 = x^Ln2

 

[soroban]

Hello, JPJ!

\(\displaystyle \text{Why can I say: }\:2^{\ln x} \:=\:x^{\ln 2}\,?\)

Never mind about it . . . just figured it out:

. . \(\displaystyle 2^{\ln x} \:=\: \left(e^{\ln2}\right)^{\ln x} \:=\:e^{\ln2\cdot\ln x} \:=\: \left(e^{\ln x}\right)^{\ln 2} \:=\: x^{\ln 2}\)

Elegant proof . . . Good work!