solve the exponential function

Question

solve $(\sqrt6-\sqrt5)^x-\sqrt6-\sqrt5=0$

I know the result is $-1$ but I don't know how to prove it. I have tried to replace $\sqrt6-\sqrt5=t$ but then I have $t^x-t+2\sqrt5=0$ and I think it is wrong way in this case.

Answer 1

$$(\sqrt{6}-\sqrt{5})^x=\sqrt{6}+\sqrt{5}$$ $$\frac{(\sqrt{6}-\sqrt{5})^x}{\sqrt{6}+\sqrt{5}}=1$$ $$\frac{(\sqrt{6}-\sqrt{5})^x}{\sqrt{6}+\sqrt{5}}\cdot\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}}=1$$ $$\frac{(\sqrt{6}-\sqrt{5})^{x+1}}{1}=1$$

Can you finish from here?

Answer 2

$$(\sqrt6-\sqrt5)^x=\sqrt6+\sqrt5=\dfrac{(\sqrt6+\sqrt5)(\sqrt6-\sqrt5)}{\sqrt6-\sqrt5}=\dfrac{6-5}{\sqrt6-\sqrt5}.$$

Answer 3

$$(\sqrt6-\sqrt5)^x-\sqrt6-\sqrt5=0$$ $$(\sqrt6-\sqrt5)^x=(\sqrt6+\sqrt5)$$ $$(\sqrt6-\sqrt5)^x=(\sqrt6+\sqrt5)\frac{\sqrt6-\sqrt5}{\sqrt6-\sqrt5}$$ $$(\sqrt6-\sqrt5)^x=\frac{1}{\sqrt6-\sqrt5}$$

$$(\sqrt6-\sqrt5)^x=(\sqrt6-\sqrt5)^{-1}$$ so $$x=-1$$