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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

RE60K
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juantheron
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2 Answers2

2

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

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

enter image description here

RE60K
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2

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:

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

medicu
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