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Thread: Solving this inequality

  1. #1
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    Solving this inequality

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

    I keep trying different things but getting stuck an suggestions?
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by fattydq View Post
    28(pi^3)/12x^2 is less than 0.00001

    I keep trying different things but getting stuck an suggestions?
    Multiply through by 12x and I will say the rhs is 10^-5 to make it easier to read.

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

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

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

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

    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

    $\displaystyle x > 100\pi \sqrt{\frac{70\pi}{3}} = \frac{100\pi \sqrt{210\pi}}{3}$
    Last edited by e^(i*pi); Apr 15th 2009 at 04:41 PM. Reason: forgot to add in sqrt tags and pi
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  3. #3
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    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?
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  4. #4
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    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 $\displaystyle \frac{\sqrt{210\pi}}{3}$
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  5. #5
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    Quote Originally Posted by e^(i*pi) View Post
    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 $\displaystyle \frac{\sqrt{210\pi}}{3}$
    I still don't understand why 10^-5 is .00001 and not .0001...
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  6. #6
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    Quote Originally Posted by fattydq View Post
    I still don't understand why 10^-5 is .00001 and not .0001...
    Scientific notation

    If you like, consider

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