Question:solve $(\sqrt{2}+1)^x +(\sqrt{2}-1)^x=6^{x/2}$
My try:First I was trying to solve it algebrically and tried some things like squaring both sides and tried to simplify but anything didn't came up.So i started graphing LHS and RHS seperatly and found intersection of graphs at x=2.So i found the solution,but this wont work always.So my questions are
Can this be solved algebrically?If yes then how? if no then why not? Is there some general way to solve all exponential equations?
On
Unless there is some sneaky way to rewrite this equation I don't think it's really possible to solve algebraically. The problem as I see it is that we have two distinct terms on the LHS. This means that we cannot easily rewrite the expressions in the LHS and RHS so that they are expressed in the same base, in a simple way (by using $e$ and $\ln$, for example).
In the given case and similar cases, numerical methods to find a good approximation (Newton-Raphson) may be needed. If you have a very messy equation (like the one you gave) it may not always be worth it trying to solve it exactly.
On
\begin{align} (\sqrt{2}+1)^x +(\sqrt{2}-1)^x&=6^{x/2} \end{align}
It looks that this equation has specially constructed constants in order to be solvable.
Let $x=2y$: \begin{align} \left((\sqrt{2}+1)^2\right)^y +\left((\sqrt{2}-1)^2\right)^y &=6^{y} \\ \left(3+2\sqrt{2}\right)^y + \left(3-2\sqrt{2}\right)^y &=6^{y} \end{align}
Note that \begin{align} 3-2\sqrt{2}&=\frac{1}{3+2\sqrt{2}}, \\ 6&=3+2\sqrt{2} +3-2\sqrt{2} \end{align}
Let $a=(\sqrt{2}+1)^2=3+2\sqrt{2}$, $a>0$. Then we have \begin{align} a^y+\frac{1}{a^y} &= \left( a+\frac{1}{a} \right)^y \\ a^{2y}+1 &= \left( a^2+1 \right)^y, \end{align} with an obvious real solution $y=1$, $x=2$.
Dividing the both sides by $6^{x/2}=(\sqrt 6)^x$ gives $$\left(\frac{\sqrt 2+1}{\sqrt 6}\right)^x+\left(\frac{\sqrt 2-1}{\sqrt 6}\right)^x=1$$ Since $0\lt \frac{\sqrt 2\pm 1}{\sqrt 6}\lt 1$, the left hand side is strictly decreasing. Hence, $x=2$ is the only solution.