The equation I need to solve is $$4^{x+\sqrt{x^2-2}}-3 \cdot 2^{x-1+\sqrt{x^2-2}}=10$$
What have I done so far
$$4^x \cdot 4^\sqrt{x^2-2}-3\cdot 2^x \cdot 2^{-1}\cdot 2^\sqrt{x^2-2}=10$$ $$4^x \cdot 4^\sqrt{x^2-2}-3\cdot 2^x \cdot \frac{1}{2}\cdot2^\sqrt{x^2-2}=10$$ $$4^x \cdot 4^\sqrt{x^2-2}-3\cdot \frac{2^x}{2} \cdot 2^\sqrt{x^2-2}=10$$ How can I continue?
Your equation can be written in the following form $$(2^{x+\sqrt{x^2-2}})^2-3/2(2^{x+\sqrt{x^2-2}})=10$$ Now substitute $$2^{x+\sqrt{x^2-2}}=t$$