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- Feb 14, 2012

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If $\log_4 (a+2b)+\log_4 (a-2b)=1$, find the minimum of $|a|-|b|$.

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- Thread starter
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- #1

- Feb 14, 2012

- 3,756

If $\log_4 (a+2b)+\log_4 (a-2b)=1$, find the minimum of $|a|-|b|$.

- Feb 13, 2012

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If $\log_4 (a+2b)+\log_4 (a-2b)=1$, find the minimum of $|a|-|b|$.

$\displaystyle (a + 2\ b)\ (a - 2\ b) = 4 -> b = \frac{\sqrt{a^{2} - 4}}{2}\ (1)$

... so that the problem is to minimize respect to a the function...

$\displaystyle f(a) = a - \frac{\sqrt{a^{2} - 4}}{2}\ (2)$

Kind regards

$\chi$ $\sigma$

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- Feb 14, 2012

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Thanks for participating,

$\displaystyle (a + 2\ b)\ (a - 2\ b) = 4 -> b = \frac{\sqrt{a^{2} - 4}}{2}\ (1)$

... so that the problem is to minimize respect to a the function...

$\displaystyle f(a) = a - \frac{\sqrt{a^{2} - 4}}{2}\ (2)$

Kind regards

$\chi$ $\sigma$

Yes, your interpretation to the problem is correct and using the calculus method to find the minimum point of the function $\displaystyle f(a) = a - \frac{\sqrt{a^{2} - 4}}{2}\ (2)$, one will get the minimum value of $|a|-|b|=\dfrac{\sqrt{15}}{2}$ when $a=\dfrac{8}{\sqrt{15}}$.