See that if $$e^x=y\ implies e^{-x} = frac{1}{y}\ implies -x = lnleft(frac{1}{y}

ight) \ implies x = -lnleft(frac{1}{y}

ight)$$

There are several ways to show this.$$ ln frac{1}{2} = ln 2^{-1} = -1 cdot ln 2 = – ln 2 ext{.} $$

$$ ln frac{1}{2} = ln 1 – ln 2 = 0 – ln 2 = – ln 2 ext{.} $$

Using your fact, suppose $ln frac{1}{2} = x$, then $frac{1}{2} = occupychristmas.orgrm{e}^x$, so $2 = frac{1}{occupychristmas.orgrm{e}^x} = occupychristmas.orgrm{e}^{-x}$. But this says, using your fact again, $ln 2 = -x$, 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$$giving $$x=-ln2$$

In general, 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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