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

Question
Logarithms
Solve $$\displaystyle{\log{{4}}}{\left({x}+{3}\right)}+{\log{{4}}}{\left({2}-{x}\right)}={1}$$

2021-01-11
$$\displaystyle{\log{{4}}}{\left({x}+{3}\right)}+{\log{{4}}}{\left({2}-{x}\right)}={1}$$
$$\displaystyle{\log{{4}}}{\left[{\left({x}+{3}\right)}\cdot{\left({2}-{x}\right)}\right]}={1}$$
$$\displaystyle{\left({x}+{3}\right)}\cdot{\left({2}-{x}\right)}={4}$$
$$\displaystyle-{x}^{{2}}-{x}+{6}={4}$$
$$\displaystyle-{x}^{{2}}-{x}+{2}={0}{\quad\text{or}\quad}{x}^{{2}}+{x}-{2}={0}$$
$$\displaystyle{\left({x}-{1}\right)}{\left({x}+{2}\right)}={0}$$

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