Real solution of exponential equation [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 $3^x+4^x+5^x = 1+x^2$
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 not
Let $f(x) = 2^x-x^2-1$ , Then $f^{'}(x) = 2^x \ln (2)-2x$ and $f^{''}(x) = 2^x\left(\ln(2)\right)^2-2$
Now we will check the Interval of $x$ in which function $f(x)$ is Increasing or decreasing.
$f^{'}(x)=2^x \cdot \ln (2)-2x >0$ for $x\leq 0$
Now I did not understand How can i proceed after that
plz help me
Thanks
 A: Problem 1: First note that there are no negative solutions. For $2^x$ is increasing in the interval $(-\infty,0)$, 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 
$$f'(x)=\ln 2-\frac{2x}{1+x^2}.$$
This is positive whenever $x^2\ln 2-2x+\ln 2$ is positive. 
The 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 $x=2.5$. 
Problem 2: For the equation $4^x=x^2$, note that $4^x$ is increasing, and $x^2$ is decreasing in the interval $(-\infty,0)$. 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 $f'(x)=\ln 2-\frac{1}{x}$. 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. 
