How to solve this hard equation?

2,822
1. Condition for the root: $0 < x \le 2$.

2. With this condition, we have:

$\begin{array}{l} 3\log _3^2\left( {\sqrt {2 + x} + \sqrt {2 - x} } \right) + 2{\log _{\frac{1}{3}}}\left( {\sqrt {2 + x} + \sqrt {2 - x} } \right).{\log _3}\left( {9{x^2}} \right) + {\left( {1 - {{\log }_{\frac{1}{3}}}x} \right)^2} = 0\\ \Leftrightarrow 3\log _3^2\left( {\sqrt {2 + x} + \sqrt {2 - x} } \right) - 4{\log _3}\left( {\sqrt {2 + x} + \sqrt {2 - x} } \right).{\log _3}\left( {3x} \right) + \log _3^2\left( {3x} \right) = 0\\ \Leftrightarrow \left[ {{{\log }_3}\left( {\sqrt {2 + x} + \sqrt {2 - x} } \right) - {{\log }_3}\left( {3x} \right)} \right].\left[ {3{{\log }_3}\left( {\sqrt {2 + x} + \sqrt {2 - x} } \right) - {{\log }_3}\left( {3x} \right)} \right] = 0 \end{array}$

1. 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.

1. In a word, this equation has a root: $x = \frac{{2\sqrt {17} }}{9}$, only.
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user355794

Updated on July 23, 2022

• 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 years
Can you show what you've tried so far?
• user355794 over 6 years
@John, I have no idea.
• John_dydx over 6 years
Also could you clarify the first term in the expression-is it really $3\log_3 2(\sqrt{2+x} + \sqrt{2-x})$?
• Zain Patel over 6 years
@John, it's the logarithm squared.
• John_dydx over 6 years
Really? In that case we better edit it to make it really clear.
• hardmath over 6 years
Note that $\log_\frac{1}{3} x = - \log_3 x$.