What is the minimum value of |a|-|b| when $\log_4 (a+2b)+\log_4 (a-2b)=1$?

[anemone]

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

 

[chisigma]

Re: Find the minimum of |a|-|b|

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

Spoiler

From the initial conditions we derive immediately...

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

... 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

$\chi$ $\sigma$

 

[anemone]

Re: Find the minimum of |a|-|b|

Spoiler

From the initial conditions we derive immediately...

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

... 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

$\chi$ $\sigma$

Thanks for participating, chisigma! I noticed you stopped half-way and probably you could eyeball the answer from where you have stopped?:p

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 $a=\dfrac{8}{\sqrt{15}}$.