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Solve the equation: $2{\log _3}x + {\log _3}({x^2} - 3) = {\log _3}0.5 + {5^{{{\log }_5}({{\log }_3}8)}}$

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Hint: Use logarithmic properties.
 ${a^{{{\log }_a}x}} = x$
$\log {m^n} = n\log m$
$\log x{\text{ is real }}\forall x > 0$

 We have given the equation $2{\log _3}x + {\log _3}({x^2} - 3) = {\log _3}0.5 + {5^{{{\log }_5}({{\log }_3}8)}}$ which can be written as ${\log _3}{x^2} + {\log _3}({x^2} - 3) = {\log _3}0.5 + {\log _3}8$ and on further solving we’ll get $2{\log _3}x + {\log _3}({x^2} - 3) = {\log _3}0.5 + {\log _3}4$ . This equation is equivalent to the system
\[
  \left\{ {\begin{array}{*{20}{c}}
  {{x^2} > 0} \\
  {{x^2} - 3 > 0} \\
  {{x^2}({x^2} - 3) = 4}
\end{array}} \right. \\
   \Rightarrow \left\{ {\begin{array}{*{20}{c}}
  {x < 0{\text{ and }}x > 0{\text{ }}} \\
  {x < \sqrt 3 {\text{ and x}} > \sqrt 3 } \\
  {({x^2} - 4)({x^2} + 1) = 0}
\end{array}} \right. \\
   \Rightarrow {x^2} - 4 = 0 \\
  \therefore x = \pm 2,{\text{ but }}x > 0 \\
\]
Consequently, $x = 2$ is only the root of the given equation.

Note: When you are using log properties, be careful with the base. When the question says “$\ln $”, it means base is e. On the other hand, when it says “log”, it means the base Is 10, else wise questions will always write base.
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