Find the sets of real solutions

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1. Jan 9, 2016

diredragon

[b[1. The problem statement, all variables and given/known data[/b]

$|4^{3x}-2^{4x+2}*3^{x+1}+20*12^x*3^x| \ge 8*6^x(8^{x-1}+6^x)$

The sets containing the real solutions for some numbers $a, b, c, d,$ such that $-\infty < a < b < c < d < +\infty$ is of the form $(-\infty, a] \cup [b, c] \cup [d, +\infty)$. Prove it by obtaining the solution set.

2. Relevant equations

3. The attempt at a solution

So the problem is asking me to solve the inequality by finding some numbers $a, b, c, d$ and find the form of the real solution set that should be in the form given above.

Here is my current work:

1) $4^{3x}-2^{4x+2}*3^{x+1}+20*12^x*3^x \geq 8*6^x(8^{x-1}+6^x)$

2) $4^{3x}-2^{4x+2}*3^{x+1}+20*12^x*3^x \leq -8*6^x(8^{x-1}+6^x)$

1)

$2^{6x}-2^{4x+2}*3^{x+1}+20*2^{2x}*3^{2x} \geq 8*3^x*2^x(2^{3x-3}+2^x*3^x)$

$2^{6x}-12*2^{4x}*3^{x}+20*2^{2x}*3^{2x} \geq 8*3^{2x}*2^{2x}+8*3^{x}*2^{4x-3}$

$2^{6x}-12*2^{4x}*3^{x}+12*2^{2x}*3^{2x} \geq 3^{x}*2^{4x}$

$2^{6x}-13*2^{4x}*3^{x}+12*2^{2x}*3^{2x} \geq 0$

$2^{4x}-13*2^{2x}*3^{x}+12*3^{2x} \geq 0$

2)

$2^{4x}-11*2^{2x}*3^{x}+28*3^{2x} \leq 0$

not sure now how to continue, am i on the right track? (latex: messed up something again, cant find what.)

Last edited by a moderator: Jan 10, 2016
2. Jan 9, 2016

haruspex

See if you can factorise that.

3. Jan 9, 2016

diredragon

I can factor out $3^x$ out of the last two. $2^{4x} - 3^x(13*2^{2x} + 3^x) \geq 0$

4. Jan 9, 2016

haruspex

Right, and do you notice a common factor of the first two? Does that suggest a way of factorising the whole?
It might help if you substitute a for 2x and b for 3x.

5. Jan 10, 2016

diredragon

$a^4 - b(13*a^2 + b) \geq 0$
Sorry, but i dont see a way to fact out the whole thing... what do $a^4$ and $b$ have in common. This kind a looks like a quadratic if i set $g= a^2$ and if $b=constant$. Other than that i dont know

6. Jan 10, 2016

haruspex

Sorry, I made two mistakes. I didn't notice some errors in your post #3, and I meant to say write a for 4x. Going back to the original form but using the a, b substitution, you have a2-13ab+12b2.
If you can't spot the factors just apply the quadratic formula.

7. Jan 10, 2016

diredragon

How can i apply the formula when there are two square terms in here?
I tried this:
$x_1=\frac{13ab + \sqrt{169a^2b^2 - 48b^2}}{2}$ is this right?

8. Jan 10, 2016

haruspex

No. You have been inconsistent regarding the role of a. In some terms you have treated it as the x in a standard quadratic, in other terms as a coefficient. Try it again.

9. Jan 10, 2016

diredragon

oops, i see. so $x_1=\frac{13b + \sqrt{169b^2 - 48b^2}}{2}$, $x_1=\frac{13b - \sqrt{169b^2 - 48b^2}}{2}$, $x_1=12b$, $x_2=b$
$(a - 12b)(a-b) \geq 0$
$(4^x - 12*3^x)(4^x-3^x) \geq 0$
i now want to solve for $=0$
$(4^x - 12*3^x)(4^x-3^x) = 0$
$(4^x - 12*3^x)=0$ or $(4^x-3^x) = 0$ from this i only see $x_1=0$ as the solution to $(4^x-3^x) = 0$ and thats because its obvious, the first one however i cant find x.

10. Jan 10, 2016

haruspex

You can make the 12 go away by merging it into the power terms. Then take logs.

11. Jan 10, 2016

Staff: Mentor

@diredragon, please don't try to mix BBCode bold tags with the LaTeX markup. They don't play well together.

I fixed your first post here, and this was the second time that I fixed the mess that you made. Please try to be more careful in the future.

12. Jan 12, 2016

diredragon

Do you mean i can make it to $4^x - 4*3^{x+1} = 0$, $4(4^{x-1} - 3{x+1}) = 0$, how now to take logs? Log the both sides? Im not sure what to do.

13. Jan 12, 2016

haruspex

I assume you meant $4(4^{x-1} - 3^{x+1}) = 0$.
Since it is zero, you can drop the factor 4. Then move one term to the other side and take logs.

14. Jan 12, 2016

diredragon

$(x-1)log_e(4) =(x+1)log_e(3)$
$xlog_e(4) - xlog_e(3)=log_e(3) + log_e(4)$
$x= \frac{log_e(3) + log_e(4)}{log_e(4) - log_e(3)}$

15. Jan 12, 2016

Bingo.