Explanation for equivalence of ln 1/2
How is it that $\ln\left(\frac{1}{2}\right) = -\ln(2)$.
If I recall correctly then $e^x=y$ is equivalent to $x = \ln(y)$.
But how does the previously mentioned equation equate?
$\endgroup$5 Answers
$\begingroup$There are several ways to show this. $$ \ln \frac{1}{2} = \ln 2^{-1} = -1 \cdot \ln 2 = - \ln 2 \text{.} $$
$$ \ln \frac{1}{2} = \ln 1 - \ln 2 = 0 - \ln 2 = - \ln 2 \text{.} $$
Using your fact, suppose $\ln \frac{1}{2} = x$, then $\frac{1}{2} = \mathrm{e}^x$, so $2 = \frac{1}{\mathrm{e}^x} = \mathrm{e}^{-x}$. But this says, using your fact again, $\ln 2 = -x$, so $x = - \ln 2$.
$\endgroup$ $\begingroup$Suppose $$x=\ln\left(\frac 1 2\right)$$ then $$e^x=\frac 1 2 = 2^{-1}$$ So $$e^{-x}=2$$ Hence, $$-x=\ln2$$ giving $$x=-\ln2$$
In general, let $z=\ln\left(y^n\right)$. We have $$ e^z=y^n$$ So $$ e^{z/n}=y$$ giving $$\frac z n=\ln y$$ Hence, $$\ln\left(y^n\right)=z=n\ln y$$
$\endgroup$ $\begingroup$See that if $$e^x=y\\ \implies e^{-x} = \frac{1}{y}\\ \implies -x = \ln\left(\frac{1}{y}\right) \\ \implies x = -\ln\left(\frac{1}{y}\right)$$
$\endgroup$ $\begingroup$$\ln(1/2)=\ln(1)-\ln(2)=-\ln(2)$
$\endgroup$ $\begingroup$Since $\ln\frac ab = \ln a - \ln b$, so $\ln\frac12 = \ln1 - \ln2 = 0 - \ln2 = -\ln 2$.
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