Find $x$ such that $5^x - sqrt{2x} -log_2{x} = 22$












0












$begingroup$


Find $x$ such that $5^x - sqrt{2x} -log_2{x} = 22$.



I have observed that the solution of this equation should be $x = 2$, I also plotted the graph of the function $f(x) = 5^x - sqrt{2x} -log_2{x}$, and it looks like an increasing function, so the solution should be unique.



However my problem is proving that this solution is unique, and showing that the function is increasing using its derivative does not seem to work, as the expression is pretty ugly.



Do you have any suggestions on how to solve this without using the derivative?






real-analysis exponential-function







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$endgroup$








  • 1




    $begingroup$
    Have you looked at the second derivative?
    $endgroup$
    – saulspatz
    Jan 6 at 23:47










  • $begingroup$
    Yes, I have, it's not any nicer.
    $endgroup$
    – Sandel
    Jan 6 at 23:48










  • $begingroup$
    it seems to me that it would always be positive, for $x$ reasonably large. Isn't that so? However, there should also be a solution near $0$ as the function goes to $infty$ as $xto0+$
    $endgroup$
    – saulspatz
    Jan 6 at 23:53


















0












$begingroup$


Find $x$ such that $5^x - sqrt{2x} -log_2{x} = 22$.



I have observed that the solution of this equation should be $x = 2$, I also plotted the graph of the function $f(x) = 5^x - sqrt{2x} -log_2{x}$, and it looks like an increasing function, so the solution should be unique.



However my problem is proving that this solution is unique, and showing that the function is increasing using its derivative does not seem to work, as the expression is pretty ugly.



Do you have any suggestions on how to solve this without using the derivative?






real-analysis exponential-function







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Have you looked at the second derivative?
    $endgroup$
    – saulspatz
    Jan 6 at 23:47










  • $begingroup$
    Yes, I have, it's not any nicer.
    $endgroup$
    – Sandel
    Jan 6 at 23:48










  • $begingroup$
    it seems to me that it would always be positive, for $x$ reasonably large. Isn't that so? However, there should also be a solution near $0$ as the function goes to $infty$ as $xto0+$
    $endgroup$
    – saulspatz
    Jan 6 at 23:53
















0












0








0





$begingroup$


Find $x$ such that $5^x - sqrt{2x} -log_2{x} = 22$.



I have observed that the solution of this equation should be $x = 2$, I also plotted the graph of the function $f(x) = 5^x - sqrt{2x} -log_2{x}$, and it looks like an increasing function, so the solution should be unique.



However my problem is proving that this solution is unique, and showing that the function is increasing using its derivative does not seem to work, as the expression is pretty ugly.



Do you have any suggestions on how to solve this without using the derivative?






real-analysis exponential-function







share|cite|improve this question









$endgroup$




Find $x$ such that $5^x - sqrt{2x} -log_2{x} = 22$.



I have observed that the solution of this equation should be $x = 2$, I also plotted the graph of the function $f(x) = 5^x - sqrt{2x} -log_2{x}$, and it looks like an increasing function, so the solution should be unique.



However my problem is proving that this solution is unique, and showing that the function is increasing using its derivative does not seem to work, as the expression is pretty ugly.



Do you have any suggestions on how to solve this without using the derivative?






real-analysis exponential-function




real-analysis exponential-function






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 6 at 23:39









SandelSandel

1765




1765








  • 1




    $begingroup$
    Have you looked at the second derivative?
    $endgroup$
    – saulspatz
    Jan 6 at 23:47










  • $begingroup$
    Yes, I have, it's not any nicer.
    $endgroup$
    – Sandel
    Jan 6 at 23:48










  • $begingroup$
    it seems to me that it would always be positive, for $x$ reasonably large. Isn't that so? However, there should also be a solution near $0$ as the function goes to $infty$ as $xto0+$
    $endgroup$
    – saulspatz
    Jan 6 at 23:53
















  • 1




    $begingroup$
    Have you looked at the second derivative?
    $endgroup$
    – saulspatz
    Jan 6 at 23:47










  • $begingroup$
    Yes, I have, it's not any nicer.
    $endgroup$
    – Sandel
    Jan 6 at 23:48










  • $begingroup$
    it seems to me that it would always be positive, for $x$ reasonably large. Isn't that so? However, there should also be a solution near $0$ as the function goes to $infty$ as $xto0+$
    $endgroup$
    – saulspatz
    Jan 6 at 23:53










1




1




$begingroup$
Have you looked at the second derivative?
$endgroup$
– saulspatz
Jan 6 at 23:47




$begingroup$
Have you looked at the second derivative?
$endgroup$
– saulspatz
Jan 6 at 23:47












$begingroup$
Yes, I have, it's not any nicer.
$endgroup$
– Sandel
Jan 6 at 23:48




$begingroup$
Yes, I have, it's not any nicer.
$endgroup$
– Sandel
Jan 6 at 23:48












$begingroup$
it seems to me that it would always be positive, for $x$ reasonably large. Isn't that so? However, there should also be a solution near $0$ as the function goes to $infty$ as $xto0+$
$endgroup$
– saulspatz
Jan 6 at 23:53






$begingroup$
it seems to me that it would always be positive, for $x$ reasonably large. Isn't that so? However, there should also be a solution near $0$ as the function goes to $infty$ as $xto0+$
$endgroup$
– saulspatz
Jan 6 at 23:53












3 Answers
3






active

oldest

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2












$begingroup$

The function is a sum of three convex functions ($5^x$, $-sqrt{2x}$ and $-log_2 x$), hence the function is itself convex. Now the limit at zero is $infty$ and $f(2)=22$. The derivative at $2$ is positive. Thus there should be exactly one more point in $]0,2[$ where the value is $22$, and no such point in $]2,infty[$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    If you mean his $f(x)=5^x-sqrt{2x}-log_2(x)$ then $f(2)=22$. But otherwise yes, this is a nice little shortcut.
    $endgroup$
    – Ben W
    Jan 7 at 0:12










  • $begingroup$
    You are right, thanks. I will fix it.
    $endgroup$
    – A. Pongrácz
    Jan 7 at 0:17



















0












$begingroup$

Actually, you should write
$$f(x)=5^x-sqrt{2x}-log_2(x)-22.$$
Now
$$f'(x)=5^xlog(5)-left(frac{sqrt{2}}{2sqrt{x}}+frac{1}{xlog(2)}right).$$
Note that
$$xmapstofrac{sqrt{2}}{2sqrt{x}}+frac{1}{xlog(2)}$$
is decreasing on $[2,infty)$, while
$$xmapsto 5^xlog(5)$$
is increasing on $[2,infty)$. Hence, $f'(x)$ is increasing on $[2,infty)$. A quick computation shows that $f'(2)>0$, and since $f'$ is increasing on $[2,infty)$ that means $f'(x)>0$ for all $[2,infty)$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    As saulspatz commented, the key point is that, since $x>0$ the second derivative is always positive.
    $$f(x)=5^x- sqrt{2x}-frac{log (x)}{log (2)}-22implies f''(x)=frac{1}{2 sqrt{2} x^{3/2}}+frac{1}{x^2 log (2)}+5^x log ^2(5)$$



    Atking into account the behavior of $log(x)$ close to $x=0^+$, there is another root. Expand $f(x)$ as a Taylor series at $x=0$ to get
    $$f(x)=-left(frac{log (x)}{log (2)}+21right)-sqrt{2} sqrt{x}+x log
    (5)+Oleft(x^{3/2}right)$$     Using the first term only
    $$frac{log (x)}{log (2)}+21=0 implies x=2^{-21}approx 4.76837times 10^{-7}$$ while solving the equation using Newton method with $x_0=2^{-21}$ would give the following iterates
    $$left(
    begin{array}{cc}
    n & x_n \
    0 & 4.768371582times 10^{-7} \
    1 & 4.765147490 times 10^{-7}\
    2 & 4.765148580times 10^{-7}
    end{array}
    right)$$






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

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      3 Answers
      3






      active

      oldest

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      active

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      active

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      2












      $begingroup$

      The function is a sum of three convex functions ($5^x$, $-sqrt{2x}$ and $-log_2 x$), hence the function is itself convex. Now the limit at zero is $infty$ and $f(2)=22$. The derivative at $2$ is positive. Thus there should be exactly one more point in $]0,2[$ where the value is $22$, and no such point in $]2,infty[$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        If you mean his $f(x)=5^x-sqrt{2x}-log_2(x)$ then $f(2)=22$. But otherwise yes, this is a nice little shortcut.
        $endgroup$
        – Ben W
        Jan 7 at 0:12










      • $begingroup$
        You are right, thanks. I will fix it.
        $endgroup$
        – A. Pongrácz
        Jan 7 at 0:17
















      2












      $begingroup$

      The function is a sum of three convex functions ($5^x$, $-sqrt{2x}$ and $-log_2 x$), hence the function is itself convex. Now the limit at zero is $infty$ and $f(2)=22$. The derivative at $2$ is positive. Thus there should be exactly one more point in $]0,2[$ where the value is $22$, and no such point in $]2,infty[$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        If you mean his $f(x)=5^x-sqrt{2x}-log_2(x)$ then $f(2)=22$. But otherwise yes, this is a nice little shortcut.
        $endgroup$
        – Ben W
        Jan 7 at 0:12










      • $begingroup$
        You are right, thanks. I will fix it.
        $endgroup$
        – A. Pongrácz
        Jan 7 at 0:17














      2












      2








      2





      $begingroup$

      The function is a sum of three convex functions ($5^x$, $-sqrt{2x}$ and $-log_2 x$), hence the function is itself convex. Now the limit at zero is $infty$ and $f(2)=22$. The derivative at $2$ is positive. Thus there should be exactly one more point in $]0,2[$ where the value is $22$, and no such point in $]2,infty[$.






      share|cite|improve this answer











      $endgroup$



      The function is a sum of three convex functions ($5^x$, $-sqrt{2x}$ and $-log_2 x$), hence the function is itself convex. Now the limit at zero is $infty$ and $f(2)=22$. The derivative at $2$ is positive. Thus there should be exactly one more point in $]0,2[$ where the value is $22$, and no such point in $]2,infty[$.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Jan 7 at 0:18

























      answered Jan 6 at 23:57









      A. PongráczA. Pongrácz

      5,9381929




      5,9381929












      • $begingroup$
        If you mean his $f(x)=5^x-sqrt{2x}-log_2(x)$ then $f(2)=22$. But otherwise yes, this is a nice little shortcut.
        $endgroup$
        – Ben W
        Jan 7 at 0:12










      • $begingroup$
        You are right, thanks. I will fix it.
        $endgroup$
        – A. Pongrácz
        Jan 7 at 0:17


















      • $begingroup$
        If you mean his $f(x)=5^x-sqrt{2x}-log_2(x)$ then $f(2)=22$. But otherwise yes, this is a nice little shortcut.
        $endgroup$
        – Ben W
        Jan 7 at 0:12










      • $begingroup$
        You are right, thanks. I will fix it.
        $endgroup$
        – A. Pongrácz
        Jan 7 at 0:17
















      $begingroup$
      If you mean his $f(x)=5^x-sqrt{2x}-log_2(x)$ then $f(2)=22$. But otherwise yes, this is a nice little shortcut.
      $endgroup$
      – Ben W
      Jan 7 at 0:12




      $begingroup$
      If you mean his $f(x)=5^x-sqrt{2x}-log_2(x)$ then $f(2)=22$. But otherwise yes, this is a nice little shortcut.
      $endgroup$
      – Ben W
      Jan 7 at 0:12












      $begingroup$
      You are right, thanks. I will fix it.
      $endgroup$
      – A. Pongrácz
      Jan 7 at 0:17




      $begingroup$
      You are right, thanks. I will fix it.
      $endgroup$
      – A. Pongrácz
      Jan 7 at 0:17











      0












      $begingroup$

      Actually, you should write
      $$f(x)=5^x-sqrt{2x}-log_2(x)-22.$$
      Now
      $$f'(x)=5^xlog(5)-left(frac{sqrt{2}}{2sqrt{x}}+frac{1}{xlog(2)}right).$$
      Note that
      $$xmapstofrac{sqrt{2}}{2sqrt{x}}+frac{1}{xlog(2)}$$
      is decreasing on $[2,infty)$, while
      $$xmapsto 5^xlog(5)$$
      is increasing on $[2,infty)$. Hence, $f'(x)$ is increasing on $[2,infty)$. A quick computation shows that $f'(2)>0$, and since $f'$ is increasing on $[2,infty)$ that means $f'(x)>0$ for all $[2,infty)$.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Actually, you should write
        $$f(x)=5^x-sqrt{2x}-log_2(x)-22.$$
        Now
        $$f'(x)=5^xlog(5)-left(frac{sqrt{2}}{2sqrt{x}}+frac{1}{xlog(2)}right).$$
        Note that
        $$xmapstofrac{sqrt{2}}{2sqrt{x}}+frac{1}{xlog(2)}$$
        is decreasing on $[2,infty)$, while
        $$xmapsto 5^xlog(5)$$
        is increasing on $[2,infty)$. Hence, $f'(x)$ is increasing on $[2,infty)$. A quick computation shows that $f'(2)>0$, and since $f'$ is increasing on $[2,infty)$ that means $f'(x)>0$ for all $[2,infty)$.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Actually, you should write
          $$f(x)=5^x-sqrt{2x}-log_2(x)-22.$$
          Now
          $$f'(x)=5^xlog(5)-left(frac{sqrt{2}}{2sqrt{x}}+frac{1}{xlog(2)}right).$$
          Note that
          $$xmapstofrac{sqrt{2}}{2sqrt{x}}+frac{1}{xlog(2)}$$
          is decreasing on $[2,infty)$, while
          $$xmapsto 5^xlog(5)$$
          is increasing on $[2,infty)$. Hence, $f'(x)$ is increasing on $[2,infty)$. A quick computation shows that $f'(2)>0$, and since $f'$ is increasing on $[2,infty)$ that means $f'(x)>0$ for all $[2,infty)$.






          share|cite|improve this answer









          $endgroup$



          Actually, you should write
          $$f(x)=5^x-sqrt{2x}-log_2(x)-22.$$
          Now
          $$f'(x)=5^xlog(5)-left(frac{sqrt{2}}{2sqrt{x}}+frac{1}{xlog(2)}right).$$
          Note that
          $$xmapstofrac{sqrt{2}}{2sqrt{x}}+frac{1}{xlog(2)}$$
          is decreasing on $[2,infty)$, while
          $$xmapsto 5^xlog(5)$$
          is increasing on $[2,infty)$. Hence, $f'(x)$ is increasing on $[2,infty)$. A quick computation shows that $f'(2)>0$, and since $f'$ is increasing on $[2,infty)$ that means $f'(x)>0$ for all $[2,infty)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 6 at 23:52









          Ben WBen W

          2,107615




          2,107615























              0












              $begingroup$

              As saulspatz commented, the key point is that, since $x>0$ the second derivative is always positive.
              $$f(x)=5^x- sqrt{2x}-frac{log (x)}{log (2)}-22implies f''(x)=frac{1}{2 sqrt{2} x^{3/2}}+frac{1}{x^2 log (2)}+5^x log ^2(5)$$



              Atking into account the behavior of $log(x)$ close to $x=0^+$, there is another root. Expand $f(x)$ as a Taylor series at $x=0$ to get
              $$f(x)=-left(frac{log (x)}{log (2)}+21right)-sqrt{2} sqrt{x}+x log
              (5)+Oleft(x^{3/2}right)$$               Using the first term only
              $$frac{log (x)}{log (2)}+21=0 implies x=2^{-21}approx 4.76837times 10^{-7}$$ while solving the equation using Newton method with $x_0=2^{-21}$ would give the following iterates
              $$left(
              begin{array}{cc}
              n & x_n \
              0 & 4.768371582times 10^{-7} \
              1 & 4.765147490 times 10^{-7}\
              2 & 4.765148580times 10^{-7}
              end{array}
              right)$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                As saulspatz commented, the key point is that, since $x>0$ the second derivative is always positive.
                $$f(x)=5^x- sqrt{2x}-frac{log (x)}{log (2)}-22implies f''(x)=frac{1}{2 sqrt{2} x^{3/2}}+frac{1}{x^2 log (2)}+5^x log ^2(5)$$



                Atking into account the behavior of $log(x)$ close to $x=0^+$, there is another root. Expand $f(x)$ as a Taylor series at $x=0$ to get
                $$f(x)=-left(frac{log (x)}{log (2)}+21right)-sqrt{2} sqrt{x}+x log
                (5)+Oleft(x^{3/2}right)$$                 Using the first term only
                $$frac{log (x)}{log (2)}+21=0 implies x=2^{-21}approx 4.76837times 10^{-7}$$ while solving the equation using Newton method with $x_0=2^{-21}$ would give the following iterates
                $$left(
                begin{array}{cc}
                n & x_n \
                0 & 4.768371582times 10^{-7} \
                1 & 4.765147490 times 10^{-7}\
                2 & 4.765148580times 10^{-7}
                end{array}
                right)$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  As saulspatz commented, the key point is that, since $x>0$ the second derivative is always positive.
                  $$f(x)=5^x- sqrt{2x}-frac{log (x)}{log (2)}-22implies f''(x)=frac{1}{2 sqrt{2} x^{3/2}}+frac{1}{x^2 log (2)}+5^x log ^2(5)$$



                  Atking into account the behavior of $log(x)$ close to $x=0^+$, there is another root. Expand $f(x)$ as a Taylor series at $x=0$ to get
                  $$f(x)=-left(frac{log (x)}{log (2)}+21right)-sqrt{2} sqrt{x}+x log
                  (5)+Oleft(x^{3/2}right)$$                   Using the first term only
                  $$frac{log (x)}{log (2)}+21=0 implies x=2^{-21}approx 4.76837times 10^{-7}$$ while solving the equation using Newton method with $x_0=2^{-21}$ would give the following iterates
                  $$left(
                  begin{array}{cc}
                  n & x_n \
                  0 & 4.768371582times 10^{-7} \
                  1 & 4.765147490 times 10^{-7}\
                  2 & 4.765148580times 10^{-7}
                  end{array}
                  right)$$






                  share|cite|improve this answer









                  $endgroup$



                  As saulspatz commented, the key point is that, since $x>0$ the second derivative is always positive.
                  $$f(x)=5^x- sqrt{2x}-frac{log (x)}{log (2)}-22implies f''(x)=frac{1}{2 sqrt{2} x^{3/2}}+frac{1}{x^2 log (2)}+5^x log ^2(5)$$



                  Atking into account the behavior of $log(x)$ close to $x=0^+$, there is another root. Expand $f(x)$ as a Taylor series at $x=0$ to get
                  $$f(x)=-left(frac{log (x)}{log (2)}+21right)-sqrt{2} sqrt{x}+x log
                  (5)+Oleft(x^{3/2}right)$$                   Using the first term only
                  $$frac{log (x)}{log (2)}+21=0 implies x=2^{-21}approx 4.76837times 10^{-7}$$ while solving the equation using Newton method with $x_0=2^{-21}$ would give the following iterates
                  $$left(
                  begin{array}{cc}
                  n & x_n \
                  0 & 4.768371582times 10^{-7} \
                  1 & 4.765147490 times 10^{-7}\
                  2 & 4.765148580times 10^{-7}
                  end{array}
                  right)$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 7 at 6:20









                  Claude LeiboviciClaude Leibovici

                  119k1157132




                  119k1157132






























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                      Partial Derivative Guidance.

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                      0 $begingroup$ I was given this tricky problem in which I'm confuse on how to solve it and I would like some help: Let $v(x,y)$ be continuously differentiable function. Let the function $u(s,t)$ be given by $u(s,t) = sin(s)*v(s-2t,t^2)$ . Compute $du/ds$ and $du/dt$ in terms of $v$ and it's partial derivatives. My attempt (which is wrong): Using chain rule: $du/ds=(du/dx)(dx/ds) + (du/dy)(dy/ds)$ $du/dt=(du/dx)(dx/dt) + (du/dx)(dy/dt)$ So, $du/ds = sin(s)*v(x,y)*s + sin(s)*v(x,y)*0$ $du/dt = 2t*sin(s)*v(x,y) - 2*sin(s)*v(x,y)$ Thanks, calculus multivariable-calculus share | cite | improve this question edited Jan 19 at 19:38...
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                      Understanding the size os this class of aleatory events

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                      0 1 $begingroup$ Hi I am having problems to understand this: I have: $Omega = [0,1]$ , $Lambda =$ all subsets with defined length Let $A_{0}$ = {A $subset$ [0,1]: A is finite union of intervals} 1º: My doubt is that $A_{0}$ is a algebra but its not a sigma algebra. Why? Also, A = $(0,1/2) cup (1/2,3/4) cup (3/4,7/8) cup ... cup(1-1/2^{n},1-1/2^{n+1})...$ 2º: So the class of aleatory events would be bigger than $A_{0}$ . Why? How the set A is greater than $A_{0}$ ? measure-theory share | cite | improve this question asked Jan 8 at 19:17 Diogo Bastos Diogo Bastos 116 6 ...
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