Originally Posted by
e^(i*pi) No, 1^x = 1 for all real x. The 10^-5 is from the decimal point being moved 5 squares to the left from 1.0. Since we use base 10 this corresponds to 10^-5
The exponent on $\displaystyle \pi$ went when I took the square root of the right hand side, using the rule
$\displaystyle \sqrt{ab} = \sqrt{a} \sqrt{b}$ and
$\displaystyle \sqrt(a^2) = a$.
Since I skipped a few steps I'll explain in more detail :]
$\displaystyle
x^2 > \frac{28 \pi^3}{12 \times 10^{-5}}
$
$\displaystyle x > \pm \sqrt{\frac{28 \pi^3}{12 \times 10^{-5}}}$
$\displaystyle x > \pm \sqrt{\frac{7}{3} \times 10^4 \times 10 \times \pi^2 \times \pi}$
Using the first rule above:
$\displaystyle x > \pm \sqrt{\frac{7}{3}} \times \sqrt{10^4} \times \sqrt{10} \times \sqrt{\pi^2} \times \sqrt{\pi}$
Note that:
$\displaystyle 10^5 = 10 \times 10^4$
$\displaystyle \pi^3 = \pi^2 \times \pi$
Using the second rule I can cancel out the $\displaystyle \sqrt{10^4}$ and $\displaystyle \sqrt{\pi^2}$
$\displaystyle x > \pm \sqrt{\frac{7}{3}} \times 100 \times \sqrt{10} \times \pi \times \sqrt{\pi}$
Recombining the square roots I have left using the first rule I get
$\displaystyle x > \pm 100\pi \times \sqrt{\frac{7}{3} \times 10 \times \pi} = 100\pi \sqrt{\frac{70\pi}{3}}$
The last step I did was rationalising the denominator - sometimes it's optional, sometimes not - depends what you covered in class. Since $\displaystyle \sqrt{\frac{70\pi}{3}} = \frac{\sqrt{70\pi}}{\sqrt{3}}$
This can be multiplied by $\displaystyle \frac{\sqrt3}{\sqrt3}$ to give $\displaystyle \frac{\sqrt3 \times \sqrt{70\pi}}{3}$
Using the rule at the top I can combine the numerators to give $\displaystyle \frac{\sqrt{210\pi}}{3}$