  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\$.

You are watching: Ln(-1/2) 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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