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Algebra inequalities and exponents

  1. Sep 4, 2012 #1
    I have been stuck on two question for sometime, and would appreciate some guidance to where I am going wrong. Here are the two questions I seem to have trouble understanding.

    1. 2x+5 < x-1/4
    I have had numerous attempts at this equation and seem to get the answer wrong each time. The book says the ans= - 3. Here a the two methods I have tried to get the correct answer:

    first attempt: 2x+5 < x-1/4 = 2x-x > -1 -5 / 4 = x > -6 / 4

    second attempt: 2x+5 < x-1/4 = 4(2x+5) < 4(x-1/4) = 8x+20 < 4x - 4/4 = 8x+20= x

    8x-x+20=0 = 8x-x= 20 which I then simplify to x = 20/7. As you can properly tell I am not getting the right answer could some please set me straight.

    2. 4^2n-3= 16
    This question I would apprectiate if someone could show how to work this out, as I seem to have no idea where to start. I would normally try and find something related to it on Google but have had no such luck. I don't really know where to look. Is there a specific name for the type of equation?
     
  2. jcsd
  3. Sep 4, 2012 #2
    2x+5<x-1/4
    x<-20/4-1/4
    x<-21/4

    4(2x+5)<4(x-1/4)
    8x -4x < -20-1
     
  4. Sep 4, 2012 #3
    Your lack of brackets or LaTeX is both annoying and confusing.

    For your first question, the answer [itex]x < -3 [/itex] only comes about if the equation is [itex]2x + 5 < \frac{x-1}{4}[/itex], so I'm assuming that's what it is. From your working, it's a bit confusing as to what equation you're trying to solve. Rewriting your working with brackets/LaTeX would make everything a lot clearer for us to understand.

    Again, with your second question, it's ambiguous. Do you mean [itex]4^{2n} - 3= 16[/itex] pr [itex]4^{2n-3} = 16[/itex]?

    If it's [itex]4^{2n-3} = 16[/itex], then that's easy to solve.

    [itex]4^{2n-3} = 16[/itex] and [itex]4^2 = 16[/itex], then [itex]2n-3=2[/itex].

    If you meant [itex]4^{2n} - 3= 16[/itex], then that's a bit more complicated, have you used logarithms before?
     
  5. Sep 4, 2012 #4
    I will give it another go, the reason why I haven't put latex in is because I am having trouble with my browser, seems to screw it up. I do apologize and appreciate the help.
     
  6. Sep 4, 2012 #5
    Here is my working out with the correct latex; hopefully.


    First attempt:
    2x + 5 < [itex]\frac{x-1}{4}[/itex] [itex]\rightarrow[/itex] x + 5 < [itex]\frac{-1}{4}[/itex] [itex]\rightarrow[/itex]

    x < [itex]\frac{-6}{4}[/itex]

    I only put the first attemp in because I think my second is completely wrong. I would like to know how this equations works out to be x = -3

    The second equation is: 42n-3= 16. I don't understand where you get the 2n-3=2. How did you come to this answer, I understand if you simplify the equation you get, 2.5 which is the answer, but I dont understand where you go the equation from. Could you explain it in a step by step.
     
    Last edited: Sep 4, 2012
  7. Sep 4, 2012 #6
    For the first equation, your first step is wrong. Where you've attempted to subtract [itex]x[/itex] from both sides, you've actually subtracted [itex]x[/itex] from the left side, but only [itex]\frac{x}{4}[/itex] from the left side. Adding, for example, 2 to a fraction, is not the same as adding 2 to the numerator of the fraction. In fact, your second attempt is almost right, except the final bit, the "[itex]=x[/itex]", I have no idea how you got to that. Redo it from [itex] 8x+20 < \frac{4x-4}{4} [/itex].

    For the second equation, you have that 4 to the power of something is 16, or: [itex]4^x = 16[/itex]. Is it not obvious from this what [itex]x[/itex] is equal to? What power do you have to put 4 to, to get 16?
     
  8. Sep 4, 2012 #7
    Right now I am complete lost. Could you show me how you would workout both problems, so I have something visual to look at. It would be big help. I still cant see where the 2n-3=2 comes from. I understand that 4^2=16 but the 2n-3=2 how dose it fit in to the equation?
     
  9. Sep 4, 2012 #8

    CAF123

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    We are not allowed to provide complete solutions. For your question, you know that [itex] 4^2 = 16. [/itex] We have [itex] 4^{2n-3} = 16. [/itex] So for what value of n will we get 2 as the exponent?
     
  10. Sep 4, 2012 #9
    Sorry I did not know. I think I get what you are saying. Am I right is working it out like this:

    4^(2n-3)=16 so 4^(2n-3)=4^2, in the way I look is the 4=4 so they cancel out and you are left with: 2n-3=2 simplified is 2.5. Am I on the right train of thought.
     
  11. Sep 4, 2012 #10

    CAF123

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    I wouldn't describe the 4's as 'cancelling' out as such. In general, if we have [tex] a^x = a^y,[/tex] then [itex] x=y.[/itex] The base here, [itex] a [/itex] must be the same on both sides of the equation.
     
  12. Sep 4, 2012 #11
    I see where you are coming from now. Thank for the help. I will give another look at the first equation and see if I can post in a better way. Once aging thanks.
     
  13. Sep 4, 2012 #12
    I have now figured out the 1st equation. I dont know how I got that mixed up, some how I did. I would like to say thanks to everyone for there input.n
     
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