Solving this inequality

**fattydq** · April 15th 2009, 04:34 PM

28(pi^3)/12x^2 is less than 0.00001

I keep trying different things but getting stuck an suggestions?

**e^(i*pi)** · April 15th 2009, 04:38 PM

Originally Posted by fattydq

28(pi^3)/12x^2 is less than 0.00001

I keep trying different things but getting stuck an suggestions?

Multiply through by 12x² and I will say the rhs is 10^-5 to make it easier to read.

$28\pi^3 < 12 \times 10^{-5} x^2$

$x^2 > \frac{28 \pi^3}{12 \times 10^{-5}}$

$x^2 > \frac{7\pi^3}{3} \times 10^5$

$x > 100\pi \sqrt{\frac{70\pi}{3}}$ OR $x > -100\pi \sqrt{\frac{70\pi}{3}}$

However, if you look at the conditions it has to be greater than both so we disregard the lower limit to get a final answer of

$x > 100\pi \sqrt{\frac{70\pi}{3}} = \frac{100\pi \sqrt{210\pi}}{3}$

**fattydq** · April 15th 2009, 04:42 PM

But wouldn't it be 1 to the -5 and not 10 to the -5? or 10^-6. and what happened to that exponent in your final step, how did it just disappear?

**e^(i*pi)** · April 15th 2009, 04:54 PM

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 $\pi$ went when I took the square root of the right hand side, using the rule
$\sqrt{ab} = \sqrt{a} \sqrt{b}$ and
$\sqrt(a^2) = a$ .

Since I skipped a few steps I'll explain in more detail :]

$ x^2 > \frac{28 \pi^3}{12 \times 10^{-5}} $

$x > \pm \sqrt{\frac{28 \pi^3}{12 \times 10^{-5}}}$

$x > \pm \sqrt{\frac{7}{3} \times 10^4 \times 10 \times \pi^2 \times \pi}$

Using the first rule above:

$x > \pm \sqrt{\frac{7}{3}} \times \sqrt{10^4} \times \sqrt{10} \times \sqrt{\pi^2} \times \sqrt{\pi}$

Note that:
$10^5 = 10 \times 10^4$

$\pi^3 = \pi^2 \times \pi$

Using the second rule I can cancel out the $\sqrt{10^4}$ and $\sqrt{\pi^2}$

$x > \pm \sqrt{\frac{7}{3}} \times 100 \times \sqrt{10} \times \pi \times \sqrt{\pi}$

Recombining the square roots I have left using the first rule I get

$x > \pm 100\pi \times \sqrt{\frac{7}{3} \times 10 \times \pi} = 100\pi \sqrt{\frac{70\pi}{3}}$

The last step I did was rationalising the denominator - sometimes it's optional, sometimes not - depends what you covered in class. Since $\sqrt{\frac{70\pi}{3}} = \frac{\sqrt{70\pi}}{\sqrt{3}}$

This can be multiplied by $\frac{\sqrt3}{\sqrt3}$ to give $\frac{\sqrt3 \times \sqrt{70\pi}}{3}$

Using the rule at the top I can combine the numerators to give $\frac{\sqrt{210\pi}}{3}$

**fattydq** · April 15th 2009, 05:04 PM

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 $\pi$ went when I took the square root of the right hand side, using the rule
$\sqrt{ab} = \sqrt{a} \sqrt{b}$ and
$\sqrt(a^2) = a$ .

Since I skipped a few steps I'll explain in more detail :]

$ x^2 > \frac{28 \pi^3}{12 \times 10^{-5}} $

$x > \pm \sqrt{\frac{28 \pi^3}{12 \times 10^{-5}}}$

$x > \pm \sqrt{\frac{7}{3} \times 10^4 \times 10 \times \pi^2 \times \pi}$

Using the first rule above:

$x > \pm \sqrt{\frac{7}{3}} \times \sqrt{10^4} \times \sqrt{10} \times \sqrt{\pi^2} \times \sqrt{\pi}$

Note that:
$10^5 = 10 \times 10^4$

$\pi^3 = \pi^2 \times \pi$

Using the second rule I can cancel out the $\sqrt{10^4}$ and $\sqrt{\pi^2}$

$x > \pm \sqrt{\frac{7}{3}} \times 100 \times \sqrt{10} \times \pi \times \sqrt{\pi}$

Recombining the square roots I have left using the first rule I get

$x > \pm 100\pi \times \sqrt{\frac{7}{3} \times 10 \times \pi} = 100\pi \sqrt{\frac{70\pi}{3}}$

The last step I did was rationalising the denominator - sometimes it's optional, sometimes not - depends what you covered in class. Since $\sqrt{\frac{70\pi}{3}} = \frac{\sqrt{70\pi}}{\sqrt{3}}$

This can be multiplied by $\frac{\sqrt3}{\sqrt3}$ to give $\frac{\sqrt3 \times \sqrt{70\pi}}{3}$

Using the rule at the top I can combine the numerators to give $\frac{\sqrt{210\pi}}{3}$

I still don't understand why 10^-5 is .00001 and not .0001...

**Reckoner** · April 15th 2009, 06:13 PM

Originally Posted by fattydq

I still don't understand why 10^-5 is .00001 and not .0001...

Scientific notation

If you like, consider

$\begin{array}{rcccl} 1&=&1/1&=&10^0\\ 0.1&=&1/10&=&10^{-1}\\ 0.01&=&1/10^2&=&10^{-2}\\ 0.001&=&1/10^3&=&10^{-3}\\ 0.0001&=&1/10^4&=&10^{-4}\\ 0.00001&=&1/10^5&=&10^{-5}\\ \end{array}$

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