If $\dfrac x{x^{1.5}}=(8x)^{-1}$ and $x>0$, then $x=\;?$

Question

If $\dfrac x{x^{1.5}}=(8x)^{-1}$ and $x>0$, then $x=\;?$

I solved this many different ways and every time I get a different answer...

Attempt #1:

$$\frac1{\sqrt x}=\frac1{8x}$$ $$8x=\sqrt x$$ $$64x^2=x$$ $$x(64x-1)=0$$ $$x=0,\;\frac1{64}$$ $$x=\frac1{64}$$

Attempt #2:

$$\frac1{\sqrt x}=\frac1{8x}$$
$$\frac{\sqrt x}x=\frac1{8x}$$
$$x=8x\sqrt x$$
$$\require{cancel}\cancel{8\sqrt x=0}\quad{8\sqrt x=1}$$
$$\cancel{x=0}\quad{\sqrt x=\frac18}$$
$$\cancel{\text{no solution}}\quad x=\frac1{64}$$

Attempt #3:

$$\require{cancel}\cancel{\frac x{x^{0.5}}=1}\quad \frac {8x}{x^{0.5}}=1$$
$$\cancel{\sqrt x=1}\quad 8x=\sqrt x$$
$$\cancel{x=1}\quad x=\frac1{64}$$

Can someone provide a solution and also explain what I did wrong in each of these attempts? Thanks.

Answer 1

Your first attempt is right but the problem says $x>0$

So, the solution will only be $\dfrac{1}{64}$

In second attempt your first three steps are correct but forth step should be- $$0=8x\sqrt{x}-x$$ $$0=x(8\sqrt{x}-1)$$

In thirdt attempt, steps should be- $$\dfrac{x}{x^{1.5}}=\dfrac{1}{8x}$$ $$\dfrac{8x}{x^{0.5}}=\dfrac{x}{x}$$ $$\dfrac{8x}{x^{0.5}}=1$$ $$8x=\sqrt{x}$$

I hope it'll help.