Can't solve for x

[T3kii]

xln(1+x^2)-1=0
i got it to: (1+x^2)^x= e
but i don't know what to do next
+the question doesn't necessarily ask to solve for x, it just asks how many solutions does this function have

 

[topsquark]

You'll have to explore the solution set. You can't solve this for x, at least not without using numerical techniques.

I'd start by graphing it. Both x and ln(.) are increasing functions. What does that suggest?

-Dan

 

[skeeter]

[MATH]f(x)=x \cdot \ln(1+x^2) -1[/MATH], a function continuous over its domain.

using the intermediate value theorem,
[MATH]f(1)= \ln(2)-1 < 0[/MATH] and [MATH]f(2) = \ln(25)-1 >0 \implies f(x) =0[/MATH] for some x-value in the interval [MATH]1 < x < 2[/MATH]
you can also use the derivative of the function to show it is strictly increasing and come to a conclusion about the number of solutions