Real solution of exponential equation1) Total no. of real solution of the equation 2 x = 1 + x 2 2) Total no. of real solution of the equation 4 x = x 2 My Try: 1) Here x = 0 and x = 1 are solution of Given equation.Now we will check for other solution which are exists or notLet f ( x ) = 2 x − x 2 − 1, Then f ′ ( x ) = 2 x ln ⁡ ( 2 ) − 2 x and f ″ ( x ) = 2 x ( ln ⁡ ( 2 ) ) 2 − 2.Now we will check the Interval of x in which function f(x) is Increasing or decreasing. f ′ ( x ) = 2 x ⋅ ln ⁡ ( 2 ) − 2 x &gt; 0Now I did not understand How can i proceed after that

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Real solution of exponential equation1) Total no. of real solution of the equation       2    x    =  1  +      x    2  2) Total no. of real solution of the equation       4    x    =      x    2  My Try: 1) Here   x  =  0 and   x  =  1 are solution of Given equation.Now we will check for other solution which are exists or notLet   f  (  x  )  =      2    x    −      x    2    −  1, Then       f                          ′              (  x  )  =      2    x    ln  ⁡  (  2  )  −  2  x and       f                          ″              (  x  )  =      2    x              (      ln      ⁡      (      2      )      )        2    −  2.Now we will check the Interval of x in which function f(x) is Increasing or decreasing.      f                          ′              (  x  )  =      2    x    ⋅  ln  ⁡  (  2  )  −  2  x  &amp;gt;  0Now I did not understand How can i proceed after that

lematayp · Accepted Answer

Step 1Problem 1: First note that there are no negative solutions. For       2    x   is increasing in the interval   (  −  ∞  ,  0  ), while the function   1  +      x    2   is decreasing. And there is a root at   x  =  0.Now I would prefer to take the ln of both sides, and consider the equation   x  ln  ⁡  2  =  ln  ⁡  (  1  +      x    2    ). So let   f  (  x  )  =  x  ln  ⁡  2  −  ln  ⁡  (  1  +      x    2    )  ..We have       f    ′    (  x  )  =  ln  ⁡  2  −            2      x              1      +              x        2              ..This is positive whenever       x    2    ln  ⁡  2  −  2  x  +  ln  ⁡  2 is positive.Step 2The roots of the quadratic are roughly 0.4 and 2.5. So the function is increasing up to about 0.4, then decreasing in roughly the interval (0.4,2.5).The function reaches 0 at   x  =  1. Then it decreases for a while, and then starts to increase. In the long run,       2    x   is much larger than   1  +      x    2  , so there is an additional root somewhere beyond   x  =  2.5.

Max Macias · Accepted Answer

Step 1Problem 2: For the equation       4    x    =      x    2  , note that       4    x   is increasing, and       x    2   is decreasing in the interval   (  −  ∞  ,  0  ). Since       4    x   is below       x    2   for large negative x, and above at   x  =  0, there is exactly one negative root.To look for positive roots, it is easier to look at the equivalent       2    x    =  x, or equivalently, taking logarithms, at   x  ln  ⁡  2  −  x.Let   f  (  x  )  =  x  ln  ⁡  2  −  ln  ⁡  x. Then       f    ′    (  x  )  =  ln  ⁡  2  −      1    x  . Thus f decreases in the interval (0, ln 2), then increases. The minimum alue of f(x) is positive, so there are no positive roots.

1) Total no. of real solution of the equation 2^x=1+x^2. 2) Total no. of real solution of the equation 4^x=x^2. 3) Total no. of real solution of the equation 2^x+3^x+4^x=x^2. 4) Total no. of real solution of the equation 3^x+4^x+5^x=1+x^2.

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