The Intermediate Value Theorem states that over a closed interval [ a , b ] for line L, that there exists a value c in that interval such that f ( c ) = L.We know both functions require x &gt; 0, however this is not a closed interval.However, I went ahead on the problem anyway. I decided to solve for x. 2 x = 10 x x 2 x = 10 x 2 l o g 2 = l o g 10 x 2 = 1 l o g 2 x = 1 l o g 2 However, when graphing these two functions I found that they both met up at ∼ 2.236, which is not equal to x = 1 l o g 2 I am very new to Calculus and the Intermediate Value theorem and would appreciate a simple explanation. Thank you.

Question

The Intermediate Value Theorem states that over a closed interval   [  a  ,  b  ] for line   L, that there exists a value   c in that interval such that   f  (  c  )  =  L.We know both functions require   x  &amp;gt;  0, however this is not a closed interval.However, I went ahead on the problem anyway. I decided to solve for x.      2    x    =      10    x    x      2    x    =  10      x    2    l  o  g  2  =  l  o  g  10      x    2    =      1          l      o      g      2        x  =            1              l        o        g        2            However, when graphing these two functions I found that they both met up at   ∼  2.236, which is not equal to   x  =            1              l        o        g        2            I am very new to Calculus and the Intermediate Value theorem and would appreciate a simple explanation. Thank you.

Phoenix Owen · Accepted Answer

You have an error in one of the steps it should be as follows:  x      2    x    =  10    ⟹    ln  ⁡  (  x      2    x    )  =  ln  ⁡  10    ⟹    ln  ⁡  x  +  x  ln  ⁡  2  =  ln  ⁡  10.To use IVT, let   f  (  x  )  =  x      2    x    −  10, now argue that it is a continuous function for   x  &amp;gt;  0. After this find two positive real numbers   a and   b such that   f  (  a  )  &amp;lt;  0 and   f  (  b  )  &amp;gt;  0. Then by IVT you can claim that there must be some point   c in the interval   [  a  ,  b  ] where   f takes the value 0. This would mean   c      2    c    =  10.If you try   a  =  1, you get   f  (  1  )  =  2  −  10  =  −  8  &amp;lt;  0, now try a value for   b.

gebuiteldmz · Accepted Answer

The difference   f  (  x  )  =      2    x    −      10    x   is continuous on   (  0  ,  ∞  ). As   f  (  1  )  =  −  8  &amp;lt;  0 and   f  (  10  )  =  1023  &amp;gt;  0, the IVT tells us that there exists at least one   c  ∈  [  1  ,  20  ] with   f  (  c  )  =  0.

The Intermediate Value Theorem states that over a closed interval [a,b] for line L, that there exists a value c in that interval such that f(c)=L. We know both functions require x>0, however this is not a closed interval. However, I went ahead on the problem anyway. I decided to solve for x.

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