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


$$17^x+9^{x^2} = 23^x+3^{x^2}$$

Clearly $0,1$ are the two roots.I would prefer rough sketching the graph:

