I've been trying to solve this problem since 2 days but I can't seem to do it.
$2^x+6^x=9^x$
The furthest that I managed to reduce it to is the following
$4.5^x-3^x=1$
but I can't proceed from here. I wanted to ask if there is a method to solve this algebraically for the solution.
Thank you
On
There is not a method to solve it algebraically.
What we can do is to show that a solution exists, indeed let consider
$$f(x)=4.5^x-3^x=1$$
which only may have solution for $x>0$ and since $f(0)=0$, $f(1)=\frac32$ and
$$f'(x)=4.5^x\log(4.5)-3^x\log 3>0$$
$f(x)$ is strictly incresing for $x>0$ and therefore by IVT exactly a solution exists for $x\in(0,1)$ which can be found by numerical methods.
As you understood, you will need numerical methods for finding the zero of the function $$f(x)=9^x-6^x-2^x$$ and, as already said in @user's answer, by the intermediate value theorem you know that the solution is $x\in(0,1)$.
If you graph $f(x)$ over the range of interest, you will notice that the plot is highly curved and this is not very good for numerical methods.
Consider instead that you look for the zero of function $$g(x)=\log(9^x)-\log(6^x+2^x)$$ Graph it to see almost a straight line and this is much better. We can use one single iteration of Newton method (or Taylor series) with $x_0=0$ and $x_0=1$ to get $$g(x)= -\log (2)+x \left(-\frac{\log (2)}{2}-\frac{\log (6)}{2}+\log (9)\right)+O\left(x^2\right)\tag1$$ and get the estimate $$x_{(0)}= \frac{2 \log (2)}{2\log (9)-\log (6)- \log (2)}\approx 0.725982$$ Similarly $$g(x)=(\log (9)-\log (8))+(x-1) \left(-\frac{\log (2)}{4}-\frac{3 \log (6)}{4}+\log (9)\right)+O\left((x-1)^2\right)\tag2$$ and get the estimate $$x_{(1)}=\frac{\log (2)+3 \log (6)-4 \log (8)}{\log (2)+3 \log (6)-4 \log (9)}\approx 0.826820$$ By Darboux theorem we knew a priori that $x_{(0)}$ is an underestimate of the solution and that $x_{(1)}$ is an overestimate of it. So, it is better to start Newton method using $x_0=x_{(0)}$ and the iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.7259824579 \\ 1 & 0.8297724614 \\ 2 & 0.8316761356 \\ 3 & 0.8316767566 \end{array} \right)$$