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Exponential equation problem,

  1. Aug 7, 2016 #1
    1. The problem statement, all variables and given/known data

    2x=1/ square root of x.

    2. Relevant equations

    None that I feel ought to be included.
    3. The attempt at a solution

    To answer the question I graphed 2^x and 1/sqrt x individually. They intersect at an x value of 0.5, so x= 0.5.
    The problem is that while the answer was correct, the working out was wrong- I did the question differently to how the teacher intended.

    I've done some playing around with the problem, moved the variables around and, for lack of better phrase, 'all sorts of stuff'. I'm not really making any headway on it, and this really ought to be a easy question (worth 2 marks in the quiz)

    I have a poor knowledge of logarithms.
    Last edited: Aug 7, 2016
  2. jcsd
  3. Aug 7, 2016 #2


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    How did you find the correct answer then? Can you show that method? How did the teacher intend it then? Also, if you post in the homework section, you should use the homework template.

    So, another question, did you learn about logarithms yet?
    Last edited: Aug 7, 2016
  4. Aug 7, 2016 #3
    Thank you for replying.

    You might have missed it, or I was unclear, but I've already explained how I found the answer.

    I set both sides of the equation individually equal to y, then using a graphics calculator (casio I think) graphed y = 2^x and y=1/sqrt x. I found the point where they intersected (0.5, 1.414) and took the x value, 0.5. Since this is the only x value that the two share, it must be the value of x that makes the statement true.

    My teacher wanted me to do it "differently". I'm not sure what he/she meant, but I've tried a few different ways of solving it, with no success. These attempts usually consist of me playing around with the variables. I'm not sure what's crippling me in this, but I'd like to stress the fact that this is supposed to be an easy question.

    I apologize for not correctly using the template. If it's necessary, I'll edit my post accordingly.
  5. Aug 7, 2016 #4


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    You can not solve the problem by isolating x. Square the equation, it becomes 4x=1/x or x=1/4x. You see at once that x<1, and x=1/41/2 is the solution.
    In cases when you do not see the solution at once, you can try an iterative procedure. Start with a trial value for x and calculate x'=0.25x. Repeat always with the new x. This equation, although slowly, converges to 0.5.
  6. Aug 7, 2016 #5
    Thank you for replying.

    I'm going to apologize again here, as I think the late hour is crippling me somewhat, but I'm at a loss as to how you found x=1/4^1/2 as the solution. I get how you found 4^x=1/x, as I simplified the statement to that point in my own attempt. I realise that at this point you've all but opened the door for me, but I'm still loitering.
  7. Aug 7, 2016 #6


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    Well, I assume you tried several things, but this is what I found:

    2^x = 1/root(x)
    <=> 2^x*x^(1/2) = 1
    <=> log(2^x*x^(1/2)) = 0
    <=> log(2^x) +log(x^(1/2)) = 0
    <=> xlog2 + 1/2logx = 0
    <=> -xlog(1/2) + 1/2logx = 0
    <=>1/2logx = xlog(1/2)
    <=> x = 1/2

    So, in fact, I used symmetry to solve this. If you have not learned about logarythms yet, ignore my post.
  8. Aug 7, 2016 #7
    Thank you for sharing this.

    Unfortunately this math is simply beyond me- though I can appreciate the effort. However, I'm now going to study logarithms. This method looks useful...
  9. Aug 7, 2016 #8


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    Well, I only used some common rules:

    log(a*b) = log(a) + log(b)
    log(a^n) = n*log(a)
  10. Aug 7, 2016 #9
    And now that you've stated what rules you've used, I can actually sort of see how they apply. Regardless, I'm not at a point in my education where I'd be able to make this connection myself. Thank you again.
  11. Aug 7, 2016 #10


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    It was by inspection. You wonder what x can be and you try some values. X must be less than 1, it is clear. You can try something simple, 1/2, for example. 41/2=√4 =2, and it is the reciprocal of 1/2! So that is the solution!
    But your graphical method was equally good.
    Math_QED solved the problem using symmetry. That is also a good method and you do not need logarithm.
    You can write the original equation in the form x1/2=(1/2)x. For x=1/2, the two sides are equivalent.
  12. Aug 7, 2016 #11
    Thank you very much, both of you, for your help. I may interrogate my teacher for further information- but I feel that now I've got a firmer handle on the situation. This is solved.
  13. Aug 7, 2016 #12

    Ray Vickson

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    The solutions given below by others works only in this special case. If you had a similar equation, but with different numbers in it, it would not be easy at all. The general case ##a^x = 1/ \sqrt{x}## for ##a > 1, a \neq 2##, involves a "non-elementary" function called the Lambert W-function. In plain words, that would mean that for a given value of ##a## you would need to use something like a graphical approach, or some other numerical scheme.
  14. Aug 7, 2016 #13
    Thank you for replying.

    The concepts you are discussing are calculus, and are far out of my humble reach. As impressive as your analysis is, it's not something I can relate to. It's certainly nothing I'd be expected to know at this stag, let alone apply.

    Maybe in a few years... It's something to keep in mind, anyhow.

    I did look up the Lambert W-function, and though this might be totally unrelated, I can recall that I did some experimenting with variables and graphing and found a form of the equation which created a graph not unlike the one associated with the W-function, only it intercepted the x axis at 0.5. Again, that might be totally unrelated.
  15. Aug 7, 2016 #14

    Ray Vickson

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    You misunderstood: I was implying that for most problems similar to yours (but with different numbers) your graphical approach would be just about the best way to obtain the solution, and that your teacher would have had to award you full marks. I was not saying you needed to understand anything in calculus. I was just pointing out that the general problem is not as easy as you might imagine.
  16. Aug 8, 2016 #15
    well if you want to find the value then prove the result you can over complicate the matter by doing the following ,
    note f(x) = lnx/x , defined and derivable on ]0;+oo[ .f'(x) = (1/x)^2 * (1 - lnx) , if x is in range ]0,1[ then f'(x)<0 , f is also continuous on ]0,1[ . f is then bijective on ]0,1[ .and since f(0.5)=-2ln2 then the only value of x that would give -2ln2 with f is 0.5 -> x=0.5
  17. Aug 8, 2016 #16
    Thank you both for your replies.

    Well it wasn't as easy as it should have been. Though I actually answered it within seconds of looking at it. My teacher simply wanted the question to test a different part of my ability. When I didn't do that, but got the right answer, I got 1 mark of a possible 2.

    Interesting approach, also simply not something I can do, nor am I expected to do, nor is something I'm taught to do. While I like to think of my math ability as above average in my small class, I'm not gifted or passionate about it. Though I might be wrong, I think using your method, alan, my teacher would have to revise a bit to make sense of that. She/he is usually teaching basic trig and quadratics. She/he leaves it to other teachers to do calculus-esque things, and since you've mentioned "defined" and "derivable" I'm assuming that you're working in that area. I may be wrong, as I'm no expert on calc.
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