No. of different real values of $x$ which satisfy $17^x+9^{x^2} = 23^x+3^{x^2}.$ Number of different real values of $x$ which satisfy $17^x+9^{x^2} = 23^x+3^{x^2}.$
$\bf{My\; Try::}$Using Hit and trial $x=0$ and $x=1$ are solution of above exponential equation. 
Now we will calculate any other solution exists or not.
If $x\geq 2\;,$ Then $17^x+9^{x^2}>9^{x^2} = (6+3)^{x^2}>6^{x^2}+3^{x^2} = (6^x)^x+3^{x^2}>23^x+3^{x^2}\;,$
bcz $(6^x>23)\; \forall x\geq 2.$
So no solution in $x\in \left[2,\infty\right)$
Now i did not understand how can i calculate in $x<0$ and $0<x<1$.
Help me, Thanks

Using derivatives, is studying functions
$ f, g: R \rightarrow R, f(x)= 9^{x^2}-3^{x^2},  g(x)=23^x-17^x$ and is found:


*

*$f$ has a minimum point in the interval $(0, 1)$ and limits to $+\infty$,$-\infty$ are equal with $+\infty$;

*$g$ has a negative minimum point and limited to $-\infty$ is $0$ and to $+\infty$ is $+\infty$.


For these reasons and noting that $f$ grows faster than $g$ infinite, it follows that graphs their only two points in common.
Conclusion: The equation has exactly two real roots.
