How to solve this hard equation?
Condition for the root: $0 < x \le 2$.
With this condition, we have:
$\begin{array}{l} 3\log _3^2\left( {\sqrt {2 + x} + \sqrt {2  x} } \right) + 2{\log _{\frac{1}{3}}}\left( {\sqrt {2 + x} + \sqrt {2  x} } \right).{\log _3}\left( {9{x^2}} \right) + {\left( {1  {{\log }_{\frac{1}{3}}}x} \right)^2} = 0\\ \Leftrightarrow 3\log _3^2\left( {\sqrt {2 + x} + \sqrt {2  x} } \right)  4{\log _3}\left( {\sqrt {2 + x} + \sqrt {2  x} } \right).{\log _3}\left( {3x} \right) + \log _3^2\left( {3x} \right) = 0\\ \Leftrightarrow \left[ {{{\log }_3}\left( {\sqrt {2 + x} + \sqrt {2  x} } \right)  {{\log }_3}\left( {3x} \right)} \right].\left[ {3{{\log }_3}\left( {\sqrt {2 + x} + \sqrt {2  x} } \right)  {{\log }_3}\left( {3x} \right)} \right] = 0 \end{array}$
 Continue to solve:
a. $\begin{array}{l}{\log _3}\left( {\sqrt {2 + x} + \sqrt {2  x} } \right)  {\log _3}\left( {3x} \right) = 0\\ \Leftrightarrow \sqrt {2 + x} + \sqrt {2  x} = 3x\\ \Leftrightarrow 4 + 2\sqrt {4  {x^2}} = 9{x^2}\\ \Leftrightarrow 2\sqrt {4  {x^2}} = 9{x^2}  4\\ \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{{x^2} \ge \frac{4}{9}}\\{81{x^4}  68{x^2}=0}\end{array}} \right.\\ \Leftrightarrow {x^2} = \frac{{68}}{{81}}\end{array}$
 Because of the condition: $0 < x \le 2$.
 There is one root: $x = \frac{{2\sqrt {17} }}{9}$.
b. $\begin{array}{l}3{\log _3}\left( {\sqrt {2 + x} + \sqrt {2  x} } \right)  {\log _3}\left( {3x} \right) = 0\\ \Leftrightarrow {\left( {\sqrt {2 + x} + \sqrt {2  x} } \right)^3} = 3x\end{array}$.
Because, $0 < x \le 2$; so, $3x \le 6$.
On the other hand, $\begin{array}{l} {\left( {\sqrt {2 + x} + \sqrt {2  x} } \right)^2} = 4 + 2\sqrt {4  {x^2}} \ge 4\\ \Rightarrow {\left( {\sqrt {2 + x} + \sqrt {2  x} } \right)^3} \ge 8\end{array}$.
In this case, there is no root.
 In a word, this equation has a root: $x = \frac{{2\sqrt {17} }}{9}$, only.
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user355794
Updated on July 23, 2022Comments

user355794 4 months
I have read this equation on a journal:
$3\left(\log _3( {\sqrt {2 + x} + \sqrt {2  x} }) \right)^2 + 2{\log _{\frac{1}{3}}}\left( {\sqrt {2 + x} + \sqrt {2  x} } \right)\cdot {\log _3}\left( {9{x^2}} \right) + {\left( {1  {{\log }_{\frac{1}{3}}}x} \right)^2} = 0$
I tried to solve it; but, no hope for me.
Can you help me to solve it?

John_dydx over 6 yearsCan you show what you've tried so far?

user355794 over 6 years@John, I have no idea.

John_dydx over 6 yearsAlso could you clarify the first term in the expressionis it really $3\log_3 2(\sqrt{2+x} + \sqrt{2x})$?

Zain Patel over 6 years@John, it's the logarithm squared.

John_dydx over 6 yearsReally? In that case we better edit it to make it really clear.

hardmath over 6 yearsNote that $\log_\frac{1}{3} x =  \log_3 x$.
