Suppose $\log_{4} (x + 2y) + \log_{4 }(x - 2y) = 1$ Find the minimum value of $|x|-|y|$

The answer is 1.732.

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This is some really weird method I came up with when solving the problem. Is there a systematic technique named after this, or is this method some ad-hoc reasoning?

Christopher Boo
- 3 years, 11 months ago

Manipulating the equality with the logarithms, we get $x^2=4+4y^2 \implies |x|=\sqrt{4+4y^2}$ .

**
First Solution
**
:
Note that the sign of y is irrelevant in the second equation as y is either squared or its absolute value is taken. Therefore it can be assumed that
$y\geq0$
. The equation to be minimized becomes
$\sqrt{4+4y^2}-y$
.
(If
$y\leq0$
, then let z=|y| which results in the equation
$\sqrt{4+4z^2}-z$
and the same solution follows.)
Differentiating this to find its absolute minimum, we have
$\frac{2y}{\sqrt{1+y^2}} - 1 = 0 \implies 4y^2 = 1+y^2 \implies y= \frac{1}{\sqrt{3}}$
. Plugging this into the second equation gives its minimum:
$\sqrt{4+4(\frac{1}{\sqrt{3}}))^2}-\frac{1}{\sqrt{3}}) = \sqrt{3}$
which is approximated to 1.732.

Note: At $y= \frac{1}{\sqrt{3}}$ , $x= \frac{4}{\sqrt{3}}$ and $x -2y = \frac{2}{\sqrt{3}}> 0$ so the $log_4(x-2y)$ in the first equation is defined and this (x,y) point does exist.

**
Alternative solution using multivariable calculus
**
:
Let
$z=|y|$
and
$t=|x|$
. This makes the equations
$t^2-4z^2=4$
and
$t-z$
.
Next, define the functions
$f(t,z) = t^2-4z^2$
and
$g(t,z) = t-z$
. This solution will use constrained optimization.
At the point that
$g(t,z)$
is maximized,
$f(t,z) = 4$
and the gradients of these functions are in the same direction:
$\nabla g(t,z) = 1\hat{\imath} -1\hat{\jmath} = \lambda(2t\hat{\imath}-8z\hat{\jmath}) = \nabla f(t,z) \implies t = 4z$
Therefore,
$4 = f(4z,z) = 16z^2 - 4z^2 = 12z^2 \implies z = \frac{1}{\sqrt{3}} \implies t = \frac{4}{\sqrt{3}}$
.
So, the minimum value of
$g(t,z)$
is
$g(\frac{4}{\sqrt{3}},\frac{1}{\sqrt{3}}) = \frac{4}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \sqrt{3}$
which is approximated to 1.732.

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Although you got it right , but you have assumed $y$ to be positive. What about the other case? Please tell me if I am wrong because I am not sure of it.

Ankit Kumar Jain
- 3 years, 11 months ago

If y<=0 then let z = |y|. We have that |x|=sqrt(4+4y^2)=sqrt(4+4z^2). Therefore |x|-|y|= sqrt(4+4z^2)-z and the same solution follows.

It can be assumed that y is nonnegative because it is either squared or its absolute value is taken so its sign doesn't matter. Likewise, it can be assumed that y is nonpositive which leads to the same result as shown in this comment.

kb e
- 3 years, 11 months ago

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But you should add that to the solution. Thanks!

Ankit Kumar Jain
- 3 years, 11 months ago

I chose LaGrange Multipliers to solve this guy. Taking the following functions in x and y:

$f(x,y) = |x| - |y|, g(x,y) = x^2 - 4y^2 = 4$

and by $grad(f) = K*grad(g),$ , we obtain:

$\frac{x}{|x|} = K \cdot 2x, -\frac{y}{|y|} = K \cdot -8y \Rightarrow 4|y| = |x| \Rightarrow x = \pm 4|y|.$

Substituting this result into g(x,y) yields:

$16|y|^2 - 4y^2 = 16[\sqrt{y^2}]^2 - 4y^2 = 16y^2 - 4*y^2 = 4 \Rightarrow y = \pm \frac{1}{\sqrt{3}}$

which in turn gives $x = \pm \frac{4}{\sqrt{3}},$ , and we have the following four critical points:

$(x, y) = (\frac{4}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}); (-\frac{4}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}).$

Thus, the minimum of $(x,y)$ computes to $\frac{4}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \boxed{\sqrt{3}}.$

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Nice one..!

gaurav sethi
- 3 years, 11 months ago

Tom, if you use Latex, your solution would be nicer.

Hana Wehbi
- 3 years, 11 months ago

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Busted aren't I, Hana! This baby was a good while before I taught myself latex. It's all made-over now.....thanks!

tom engelsman
- 3 years, 11 months ago

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Thanks, l didn't wanna tell you l was having hard time reading your solution, LoL. Nice solution anyhow.

Hana Wehbi
- 3 years, 11 months ago

We can parameterize x and y by the following method

x = 2sec(theta) y = tan(theta)

Now, the problem turns into finding the minimum value of a certain trigonometric function, which can be done easily......!!

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since x-2y>0 and x+2y>0 there adding above we get x>0,x>2y

solving log we get (x^2) -4(y^2)=4 now x and y lies on that part of curve for which x>0 the curve is symmetric above and below x axis as it involve even powers of x and y. ----> y can have two values for which |x|-|y| is minimum ,taking positive y, we get to find minimum of x-y or |x-y|or |x-y|/sq.root{2} i.e. minimum distance between curve (x^2) -4(y^2)=4 and line x=y which is possible when x and y lie on the curve and common normal. ----->slope of normal at that point=135 degree -----> slope of tangent=45 degree slope of any point slope(dy/dx)=x/4y=1 ---->point must satisfy x=4y and (x^2) -4(y^2)=4 --->therefore , x=4/sq.root{3} y=1/sq.root{3} ---->x-y=sq.root{3}=1.7321

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you got 1.732 . But I got exact 1.75. Poor solution

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lol! . then please post your "1.75" and "EXCELLENT" solution .

vineet golcha
- 5 years, 7 months ago

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The equation can be simplified to

$\begin{aligned} x^2 - 4y^2 & = 4 \\ \frac{x^2}{4} - y^2 &= 1 \end{aligned}$

This is an equation of the hyperbola

Since the four quadrants are a combination of reflection or symmetry to each other, a point in one quadrant gives the same $|x|-|y|$ value as the corresponding point in the other quadrants. Hence, we only have to consider the first quadrant and can conveniently remove the absolute function notation.

The leftmost point in the first quadrant gives $x-y=2$ . Subsequently, we have $x>y$ from the logarithm constraint, thus $x-y>0$ can never be negative.

Let $m$ be the gradient of the tangent line.

Notice that $m$ is monotonically decreasing while $x$ increases. So, we would want to move $x$ to the right until it reaches $m=1$ , in which $x-y$ will be minimised.

Differentiate the function we have

$\begin{aligned} \frac{x^2}{4} - y^2 &= 1 \\ \frac{x}{2} - 2yy' &= 0 \\ x &= 4y \end{aligned}$

The minimum value is achieved when $x=4y$ , plug this into the original function we have

$\begin{aligned} \frac{x^2}{4} - y^2 &= 1 \\ \frac{16y^2}{4} - y^2 &= 1 \\ y &= \frac{1}{\sqrt{3}} \end{aligned}$

and $x=\frac{4}{\sqrt{3}}$ gives $x-y = \frac{3}{\sqrt{3}} = \sqrt{3}$ .