Find the sets of real solutions

  • #1
diredragon
323
15
[b[1. Homework Statement [/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.

Homework Equations



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, can't find what.)
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
diredragon said:
##2^{4x}-13*2^{2x}*3^{x}+12*3^{2x} \geq 0 ##
See if you can factorise that.
 
  • #3
haruspex said:
See if you can factorise that.
I can factor out ##3^x## out of the last two. ##2^{4x} - 3^x(13*2^{2x} + 3^x) \geq 0##
 
  • #4
diredragon said:
I can factor out ##3^x## out of the last two. ##2^{4x} - 3^x(13*2^{2x} + 3^x) \geq 0##
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
haruspex said:
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.
##a^4 - b(13*a^2 + b) \geq 0##
Sorry, but i don't 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 don't know
 
  • #6
diredragon said:
##a^4 - b(13*a^2 + b) \geq 0##
Sorry, but i don't 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 don't know
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
haruspex said:
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.
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
diredragon said:
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?
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.
 
  • Like
Likes diredragon
  • #9
haruspex said:
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.
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 that's because its obvious, the first one however i can't find x.
 
  • #10
diredragon said:
##(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 that's because its obvious, the first one however i can't find x.
You can make the 12 go away by merging it into the power terms. Then take logs.
 
  • #11
diredragon said:
(latex: messed up something again, can't find what.)
@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
haruspex said:
You can make the 12 go away by merging it into the power terms. Then take logs.
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? I am not sure what to do.
 
  • #13
diredragon said:
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? I am not sure what to do.
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
haruspex said:
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.
##(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
diredragon said:
##(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)}##
Bingo.
 

Similar threads

Back
Top