Discover the form of real solution set

[diredragon]

Homework Statement

$|4^{3x}-2^{4x+2}*3^{x+1}+20*12^x*3^x| > 8*6^x(8^{x-1}+6^x)$
For some numbers $a, b, c, d$ such that $-\infty < a <b < c <d < +\infty$ the real solution set to the given inequality is of the form $(-\infty, a] \cup [b, c] \cup [d, +\infty)$ Prove it by arriving at the given result.

Homework Equations

3. The Attempt at a Solution [/B]
$a) 2^{6x}-12*2^{4x}*3^x + 20*2^{2x}*3^{2x} < -3^x*2^{4x} - 8*3^{2x}*2^{2x}$
$a) 2^{6x}-11*2^{4x}*3^x + 28*2^{2x}*3^{2x} < 0$
$a) 2^{4x}-11*2^{2x}*3^x + 28*3^{2x} < 0$

$b) 2^{6x}-12*2^{4x}*3^x + 20*2^{2x}*3^{2x} > +3^x*2^{4x} + 8*3^{2x}*2^{2x}$
$b) 2^{6x}-15*2^{4x}*3^x + 12*2^{2x}*3^{2x} > 0$
$b) 2^{4x}-15*2^{2x}*3^x + 12*3^{2x} > 0$
I don't know where to go on from this. I need to get certain numbers $a, b, c, d$ so that they would fit the result.

 

[Ray Vickson]

Homework Statement

$|4^{3x}-2^{4x+2}*3^{x+1}+20*12^x*3^x| > 8*6^x(8^{x-1}+6^x)$
For some numbers $a, b, c, d$ such that $-\infty < a <b < c <d < +\infty$ the real solution set to the given inequality is of the form $(-\infty, a] \cup [b, c] \cup [d, +\infty)$ Prove it by arriving at the given result.

Homework Equations

3. The Attempt at a Solution [/B]
$a) 2^{6x}-12*2^{4x}*3^x + 20*2^{2x}*3^{2x} < -3^x*2^{4x} - 8*3^{2x}*2^{2x}$
$a) 2^{6x}-11*2^{4x}*3^x + 28*2^{2x}*3^{2x} < 0$
$a) 2^{4x}-11*2^{2x}*3^x + 28*3^{2x} < 0$

$b) 2^{6x}-12*2^{4x}*3^x + 20*2^{2x}*3^{2x} > +3^x*2^{4x} + 8*3^{2x}*2^{2x}$
$b) 2^{6x}-15*2^{4x}*3^x + 12*2^{2x}*3^{2x} > 0$
$b) 2^{4x}-15*2^{2x}*3^x + 12*3^{2x} > 0$
I don't know where to go on from this. I need to get certain numbers $a, b, c, d$ so that they would fit the result.

You can plot both sides of the inequality, and if you do that you will see that the proposed answer is incorrect: the solution set is not of the form specified.

 

[diredragon]

You can plot both sides of the inequality, and if you do that you will see that the proposed answer is incorrect: the solution set is not of the form specified.

I'm not sure about this. It certanly must be the answer as the book from which the problem was takrn suggests. It also appeared in the math test I've taken last week and the professor is going to explain the solution next week. I tried to solve it beforehand but i arrive at this last line. I can't figure out what to do next, or if I've made a mistake in betweed.

 

[Ray Vickson]

I'm not sure about this. It certanly must be the answer as the book from which the problem was takrn suggests. It also appeared in the math test I've taken last week and the professor is going to explain the solution next week. I tried to solve it beforehand but i arrive at this last line. I can't figure out what to do next, or if I've made a mistake in betweed.

My apologies: plotting is tricky, because of the vast order-of-magnitude changes in both sides over reasonable ranges of x. However, by simplifying and dividing by a common (always positive) factor, we have more easily-plotted functions, and your claimed result does turn out to be true.

 

[SammyS]

Notice that $\ 3^{2x}=(3^x)^2\$ and $\ 2^{4x}=(2^{2x})^2\$ .

 

[HallsofIvy]

The simplest way to solve an ineqVuality is often to solve the corresponding equation. Since both sides are continuous functions of x and y, the curve where 'A= B' separates 'A< B' from 'A> B'. That means you want to solve 2^{4x}-11*2^{2x}*3^x + 28*3^{2x}= 0. To do that let u= 2^x and y= 3^x. As SammyS pointed out, 2^{4x}= (2^x)^2= u^2 and 3^{2x}= (3^x)^2= y^2 so the equation becomes x^2- 11xy+ 28y^2= 0. You should find that quadratic easy to factor.

 

[diredragon]

The simplest way to solve an ineqVuality is often to solve the corresponding equation. Since both sides are continuous functions of x and y, the curve where 'A= B' separates 'A< B' from 'A> B'. That means you want to solve 2^{4x}-11*2^{2x}*3^x + 28*3^{2x}= 0. To do that let u= 2^x and y= 3^x. As SammyS pointed out, 2^{4x}= (2^x)^2= u^2 and 3^{2x}= (3^x)^2= y^2 so the equation becomes x^2- 11xy+ 28y^2= 0. You should find that quadratic easy to factor.

You mean $u=2^{2x}$ and then the equation stands as $u^2 - 11uy + 28y^2=0$.
$(u - \frac{11y}{2})^2 - \frac{9y^2}{4} = 0$
$(u - \frac{11y}{2})^2 = \frac{9y^2}{4}$
$u - \frac{11y}{2} = \frac{3y}{2}$
Is this right?

 

[Ray Vickson]

You mean $u=2^{2x}$ and then the equation stands as $u^2 - 11uy + 28y^2=0$.
$(u - \frac{11y}{2})^2 - \frac{9y^2}{4} = 0$
$(u - \frac{11y}{2})^2 = \frac{9y^2}{4}$
$u - \frac{11y}{2} = \frac{3y}{2}$
Is this right?

I think it is more complicated than that. If $f(x) = 4^{3x}-2^{4x+2}\,3^{x+1}+20 \,12^x \,3^x$ and $g(x) = 8\,6^x(8^{x-1}+6^x)$ you need to look at the two different problems $f(x) > 0\: \& \:f(x) > g(x)$ and $f(x) < 0\: \& \: -f(x) > g(x)$.

 

[diredragon]

I think it is more complicated than that. If $f(x) = 4^{3x}-2^{4x+2}\,3^{x+1}+20 \,12^x \,3^x$ and $g(x) = 8\,6^x(8^{x-1}+6^x)$ you need to look at the two different problems $f(x) > 0\: \& \:f(x) > g(x)$ and $f(x) < 0\: \& \: -f(x) > g(x)$.

The expression i got in post #7 is taken as the first part of the problem, mainly $f(x) < g(x)$ , i then needed to set $f(x) = g(x)$ but i arrive at what i posted in #7. Dont know how to continue...

 

[SammyS]

You mean $u=2^{2x}$ and then the equation stands as $u^2 - 11uy + 28y^2=0$.
$(u - \frac{11y}{2})^2 - \frac{9y^2}{4} = 0$
$(u - \frac{11y}{2})^2 = \frac{9y^2}{4}$
$u - \frac{11y}{2} = \frac{3y}{2}$
Is this right?

That can be solved for u. However there is one more solution, which you have dropped somehow.

... but ... Getting those two solutions is not that complicated.

As Halls said, factor the left hand side of

$u^2 - 11uy + 28y^2=0$​

$(u-4y)(u+7y)=0$

 

[diredragon]

That can be solved for u. However there is one more solution, which you have dropped somehow.

... but ... Getting those two solutions is not that complicated.

As Halls said, factor the left hand side of

$u^2 - 11uy + 28y^2=0$​

$(u-4y)(u+7y)=0$

So, $(u-4y)(u+7y)=0$
$u-4y=0$ and $u+7y=0$
$2^{2x-2} = 3^x$ and $2^{2x} = -7*3^{x}$
How to solve this?

 

[Mark44]

So, $(u-4y)(u+7y)=0$
$u-4y=0$ and $u+7y=0$
$2^{2x-2} = 3^x$ and $2^{2x} = -7*3^{x}$
How to solve this?

In both equations, you need to get the same base on each side of the equation. For example, $2 = e^{\ln 2}$. Use this idea and the properties of exponents to solve each equation.

 

[Ray Vickson]

So, $(u-4y)(u+7y)=0$
$u-4y=0$ and $u+7y=0$
$2^{2x-2} = 3^x$ and $2^{2x} = -7*3^{x}$
How to solve this?

Your problem has the form $|L(x)| > R(x)$, where
L(x) = 4^{3x}-2^{4x+2}*3^{x+1}+20*12^x*3^x = 64^x-12\,48^x+20\,36^x
and
R(x) = 8*6^x(8^{x-1}+6^x) = 48^x+8 \, 36^x
Write $L(x) = 36^x \, f(x)$ and $R(x) = 36^x \, g(x)$, where
f(x) = 20 - 12\,(48/36)^x + (64/36)^x = 20 - 12\,(4/3)^x + (16/9)^x
and
g(x) = 8+(48/36)^x = 8 + (4/3)^x
Since $36^x > 0$ always, the inequalities $|L(x)| > R(x)$ and $|f(x)| > g(x)$ are equivalent. Setting $(4/3)^x = y$ we have $f(x) = 20 - 12 y + y^2$ and $g(x) = 8+y$, so the problem becomes $|20 - 12 y + y^2| > y+8$.

 

[SammyS]

So, $(u-4y)(u+7y)=0$
$u-4y=0$ and $u+7y=0$
$2^{2x-2} = 3^x$ and $2^{2x} = -7*3^{x}$
How to solve this?

First of all, somebody made a mistake in factoring $\ u^2 - 11uy + 28y^2\,.\$

Of course that is: $\ (u-4y)(u-7y)\$

I think you're much better off to think of $\ 2^{2x}\$ as $\ 4^{x}\$.

Solve for u/y or y/u

Notice that $\displaystyle\ \frac u y = \left(\frac 4 3\right)^x \$

This all is handy if you are following the advice from Halls.