+0

# equation

0
168
2

Solve using power rules:

Apr 12, 2022

#1
+2522
0

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
+118196
+1

Thanks Builderboi,

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