Find the value of the variable.

Hey there, Hana.....here's a solution for yah:

Let $u = 2^{x}, v = 3^{x}$ such that we arrive at the transformation:

$2^x + 3^x + 6^x - 4^x - 9^x = u + v + uv - u^2 - v^2 = 1$ (i)

Solving the quadratic (i) for $u$ yields: $0 = u^2 + (1+v)u + (v^2 - v + 1) \Rightarrow u = \frac{-(1+v) \pm \sqrt{(1+v)^2 - 4(1)(v^2 - v + 1)}}{2} = \frac{-(1+v) \pm \sqrt{-3v^2 + 6v - 3}}{2} = \frac{-(1+v) \pm \sqrt{-3(v-1)^2}}{2}$ (ii).

The only way (ii) can yield a solution $u \in \mathbb{R}$ occurs iff $v = 3^x = 1 \Rightarrow \boxed{x = 0}.$

$2^x + 3^x + 6^x - 4^x - 9^x = 1$ rearranges to $(2^x - 3^x)^2 + (2^x - 1)(3^x - 1) = 0$ .

When $x > 0$ , both $2^x > 1$ and $3^x > 1$ , so $(2^x - 3^x)^2 + (2^x - 1)(3^x - 1) > 0$ ; and when $x < 0$ , both $2^x < 1$ and $3^x < 1$ , so $(2^x - 3^x)^2 + (2^x - 1)(3^x - 1) > 0$ .

Therefore, $x = 0$ can be the only solution to the given equation.

Note that $(3^x-1)(2^x-1) \ge 0$ for all real $x$ , with equality precisely when $x=0$ . Thus $6^x + 1 \; \ge \; 2^x+ 3^x$ for all real $x$ , with equality precisely when $x=0$ . In addition $\tfrac12(4^x + 9^x) \; \ge \; \sqrt{4^x \times 9^x} \; = \; 6^x$ for all real $x$ , with equality precisely when $4^x=9^x$ , namely when $x=0$ . Thus $4^x + 9^x + 1 \; \ge \; 2 \times 6^x + 1 \; =\; 6^x + (6^x + 1) \; \ge \; 6^x + 3^x + 2^x$ for all real $x$ , with equality precisely when $x=0$ . Thus the only solution to the equation $2^x + 3^x + 6^x - 4^x - 9^x \; = \; 1$ is $x=\boxed{0}$ .

I am going to rewrite the equation $2^x+3^x+6^x-4^x-9^x=1$ as:

$1+4^x+9^x-2^x-3^x-6^x=0$ identified as equation $A$

$\implies 1+(2^2)^x+(3^2)^x-2^x-3^x-6^x=0$

let $p= 2^x \text{ and }q=3^x$ then $A$ will become:

$\implies 1+p^2+q^2-p-q-pq=0$

$\implies 2+2p^2+2q^2-2p-2q-2pq=0$

$\implies \textcolor{#D61F06}1+\textcolor{#3D99F6}1+\textcolor{#D61F06}{p^2}+p^2+\textcolor{#3D99F6}{q^2}+q^2\textcolor{#D61F06}{-2p}\textcolor{#3D99F6}{-2q}-2pq=0$

$\implies \textcolor{#D61F06}{(1-p)^2}+\textcolor{#3D99F6}{(1-q)^2}+(p-q)^2=0$

$\implies 1-p=1-q=p-q=0 \implies p=q=1 \implies 2^x=3^x=1$

The only solution here that satisfies $2^x=3^x=1$ is $\boxed {x=0}$