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Solve using power rules:

 Apr 12, 2022
 #1
avatar+2432 
+1

SImplify as follows:

 \(\sqrt{{3^2 \over 5^2}^{-3}} - {\sqrt{x} \over 3^3} = {11^2 \over 3^3}\)

\({\sqrt {1 \over{3^2 \over 5^2}^3}} - {\sqrt{x} \over 3^3} = {11^2 \over 3^3}\)

\({\sqrt {1 \over{3^6 \over 5^6}}} - {\sqrt{x} \over 3^3} = {11^2 \over 3^3}\)

\({\sqrt {5^6 \over 3^6}} - {\sqrt{x} \over 3^3} = {11^2 \over 3^3}\)

\({5^3 \over 3^3} - {\sqrt{x} \over 3^3} = {11^2 \over 3^3}\)

\(5^3 - \sqrt{x} = 11^2\)

\(125 - \sqrt{x} = 121\)

\(\sqrt{x} = 4\)

\(\color{brown}\boxed{x = 16}\)

 Apr 13, 2022
 #2
avatar+117819 
+1

Thanks Builderboi,   cool

 

Here ia another approach. 

 

\(\left(\frac{9}{25}\right)^{-\frac{3}{2}}-\frac{\sqrt x}{27}=\left(\frac{3^3}{11^2}\right )^{-1}\\~\\ \left(\frac{3}{5}\right)^{-3}-\frac{\sqrt x}{27}=\left(\frac{27}{121}\right )^{-1}\\~\\ \left(\frac{5}{3}\right)^{3}-\frac{\sqrt x}{27}=\frac{121}{27}\\~\\ \frac{125}{27}-\frac{\sqrt x}{27}=\frac{121}{27}\\~\\ 125-\sqrt x = 121\\~\\ \sqrt x = 4\\~\\ x=16 \)

 Apr 13, 2022

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