Question

Solve $8x^{\frac{3}{2n}}-8x^{-\frac{3}{2n}}=63$

• A. $2^{2n}$
• B. $\frac{1}{2^{2n}}$
• C. $2^{3n}$
• D. $\frac{1}{2^{3n}}$

Answer 1

Answer: (A), (B)

$2^{2n}$
$\frac{1}{2^{2n}}$

Given, $8x^{\frac{3}{2n}}-8x^{-\frac{3}{2n}}=63$

$\Rightarrow 8x^{\frac{3}{2n}}-\frac{1}{8x^{\frac{3}{2n}}}=63$

Put $x^{\frac{3}{2n}}=t$, we get

$8t-\frac{8}{t}=63\Rightarrow 8t^2-8=63t$

$\Rightarrow 8t^2-63t-8=0$

$\Rightarrow 8t^2-64t+t-8=0$

$\Rightarrow 8t(t-8)+1(t-8)=0$

$\Rightarrow (8t+1)(t-8)=0$

$\Rightarrow t=-\frac{1}{8},8$

$\Rightarrow x^{\frac{3}{2n}}=t$

$\Rightarrow x=t^{\frac{2n}{3}}$

For $t=-\frac{1}{8}$, we have

$x={(-\frac{1}{8})}^{\frac{2n}{3}}={\left({(-\frac{1}{2})}^3\right)}^{\frac{2n}{3}}x={(-\frac{1}{2})}^{2n}=\frac{1}{2^{2n}}$

For $t=8$, we have

$x=8^{\frac{2n}{3}}={\left({\left(2\right)}^3\right)}^{\frac{2n}{3}}x=2^{2n}$

So, the values of $x$ are $\frac{1}{2^{2n}}$ and $2^{2n}$.