Simplify $\left(\dfrac{2^{x-1}\sqrt{2}}{2^x}\right)^2$. Answer ... If $27m^{3x}=1$ and $m>0$, find $m^{2x}+m^{-x}$. Answer ...

1. Simplify the fraction inside the brackets using index laws.

$\frac{2^{x-1}\sqrt{2}}{2^x}=2^{x-1-x}\sqrt{2}=2^{-1}\sqrt{2}=\frac{\sqrt{2}}{2}$

2. Square the simplified expression.

$\left(\frac{\sqrt{2}}{2}\right)^2=\frac{(\sqrt{2})^2}{2^2}=\frac{2}{4}=\frac{1}{2}$

1. Find the value of $m^x$ from the given equation.

$27m^{3x}=1$

$27(m^x)^3=1$

$(m^x)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3$

Since $m>0$, $m^x=\dfrac{1}{3}$.

2. Substitute $m^x=\dfrac{1}{3}$ into the expression and evaluate.

$m^{2x}+m^{-x}=(m^x)^2+\dfrac{1}{m^x}$

$=\left(\dfrac{1}{3}\right)^2+\dfrac{1}{\frac{1}{3}}$

$=\dfrac{1}{9}+3=\dfrac{28}{9}$