Show that $x^2-\ln(x+1) = 1$ has at least one solution

Question

How would one go about solving this?

My math teacher told me that i should find the solution for each of the joints(not what it's called in english?) so basically $x^2 = 1$ which has two solutions $x = 1$ and $x = -1$ then find the solution for $-\ln(x+1) = 1$ which would be $1/e-1$, but what comes after this?

Is there a formula for solving these or an equation or something?

Thanks.

Answer 1

Define

$$f(x)=x^2-\log(x+1)-1\implies f(0)=-1<0\;,\;\;f(2)=4-\log3-1>0$$

and thus by the Intermediate Value Theorem for continuous functions, there exists $\;c\in(0,2)\;$ s.t. $\;f(c)=0\;$

Answer 2

Let $f(x) = x^2 - ln(x+1) - 1$. We have

$$f(0) = -1 < 0$$ and $$f(e-1) = (e-1)^2 - ln(e) - 1 = (e-1)^2 - 2 >0$$

By Intermediate Value Theorem, there exists at least one root in $[0, e-1]$

Answer 3

Let $f(x) = x^2-\ln(x+1)$ Then $f(0)=0, f(100)=10000 -\ln 101 >1$ Now apply IVT

Answer 4

On the interval $I=(-1,+\infty) $ $$ f(x)=x^2-1-\log(x+1) $$ is a convex function fulfilling $$ \lim_{x\to -1^+}f(x)=\lim_{x\to +\infty}f(x)=+\infty,\qquad f(1)<0 $$ hence $f(x)$ has exactly two distinct zeroes in $I$.
By solving $f'(x)=0$, we also have that the absolute minimum of $f(x)$ over $I$ occurs at $x=\frac{\sqrt{3}-1}{2}$, hence we have exactly one root on the right of such a point and exactly one root on the left of such a point.