# Solve log4(x+3)+log4(2-x)=1

Solve $\mathrm{log}4\left(x+3\right)+\mathrm{log}4\left(2-x\right)=1$
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Arnold Odonnell
$\mathrm{log}4\left(x+3\right)+\mathrm{log}4\left(2-x\right)=1$
$\mathrm{log}4\left[\left(x+3\right)\cdot \left(2-x\right)\right]=1$
$\left(x+3\right)\cdot \left(2-x\right)=4$
$-{x}^{2}-x+6=4$
$-{x}^{2}-x+2=0\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}{x}^{2}+x-2=0$
$\left(x-1\right)\left(x+2\right)=0$