# Discover the form of real solution set

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1. Feb 26, 2016

### diredragon

1. The problem statement, all variables and given/known data
$|4^{3x}-2^{4x+2}*3^{x+1}+20*12^x*3^x| > 8*6^x(8^{x-1}+6^x)$
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.
2. Relevant equations
3. The attempt at a solution

$a) 2^{6x}-12*2^{4x}*3^x + 20*2^{2x}*3^{2x} < -3^x*2^{4x} - 8*3^{2x}*2^{2x}$
$a) 2^{6x}-11*2^{4x}*3^x + 28*2^{2x}*3^{2x} < 0$
$a) 2^{4x}-11*2^{2x}*3^x + 28*3^{2x} < 0$

$b) 2^{6x}-12*2^{4x}*3^x + 20*2^{2x}*3^{2x} > +3^x*2^{4x} + 8*3^{2x}*2^{2x}$
$b) 2^{6x}-15*2^{4x}*3^x + 12*2^{2x}*3^{2x} > 0$
$b) 2^{4x}-15*2^{2x}*3^x + 12*3^{2x} > 0$
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.

2. Feb 26, 2016

### Ray Vickson

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.

3. Feb 26, 2016

### diredragon

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 gonna 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.

4. Feb 26, 2016

### Ray Vickson

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.

Last edited: Feb 26, 2016
5. Feb 26, 2016

### SammyS

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

6. Feb 27, 2016

### HallsofIvy

Staff Emeritus
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.

7. Feb 28, 2016

### diredragon

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?

8. Feb 28, 2016

### Ray Vickson

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 $f(x) < 0\: \& \: -f(x) > g(x)$.

9. Feb 28, 2016

### diredragon

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...

10. Feb 28, 2016

### SammyS

Staff Emeritus
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
$u^2 - 11uy + 28y^2=0$​

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

Last edited: Feb 28, 2016
11. Feb 28, 2016

### diredragon

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?

12. Feb 28, 2016

### Staff: Mentor

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.

13. Feb 28, 2016

### Ray Vickson

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 $|20 - 12 y + y^2| > y+8$.

14. Feb 28, 2016

### SammyS

Staff Emeritus
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.

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