What is a proof that $1/2 \ln x = \ln(\sqrt x)$ where $x>0$?
Thank you.
$\endgroup$2 Answers
$\begingroup$$a\ln(x)=\ln(x^a)$ why ?
if $\ln(x)=b$ and $\ln(x^a)=c,x^a=e^c,x=e^b$ so $e^c=e^{ab}$so $c=ab$ $$\ln(x^a)=ba=a\ln(x)$$
$\endgroup$ 1 $\begingroup$$$ ln x = \int_1^x \frac{dt}{t} $$ (set $t=q^2$) $$ = \int_1^{\sqrt{x}} \frac{2qdq}{q^2} = 2 \int_1^{\sqrt{x}} \frac{dq}{q} = 2 ln \sqrt{x} $$
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