See that if $$e^x=y\ implies e^-x = frac1y\ indicates -x = lnleft(frac1y ight) \ implies x = -lnleft(frac1y ight)$$
Tright here are several means to display this.$$ ln frac12 = ln 2^-1 = -1 cdot ln 2 = - ln 2 ext. $$
$$ ln frac12 = ln 1 - ln 2 = 0 - ln 2 = - ln 2 ext. $$
Using your reality, expect $ln frac12 = x$, then $frac12 = occupychristmas.orgrme^x$, so $2 = frac1occupychristmas.orgrme^x = occupychristmas.orgrme^-x$. But this claims, utilizing your fact aobtain, $ln 2 = -x$, so $x = - ln 2$.
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Suppose $$x=lnleft(frac 1 2 ight)$$then $$e^x=frac 1 2 = 2^-1$$So$$e^-x=2$$Hence, $$-x=ln2$$providing $$x=-ln2$$
In basic, let $z=lnleft(y^n ight)$.We have$$ e^z=y^n$$So$$ e^z/n=y$$giving$$frac z n=ln y$$Hence,$$lnleft(y^n ight)=z=nln y$$
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