Question

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

Answer 1

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