1. ## complicated equation

need help with this one :/

2^(3-x) = 2^x - 2

2^(3-x) - 2^x + 2 = 0 = 2^-x * 2 * 2 * 2 - 2^x + 2

someone who can help with the rest ?

2. $\displaystyle 2^{3-x}=2^x-2\;\Leftrightarrow\;\dfrac{2^3}{2^x}=2^x-2$

Now, use the substitution $\displaystyle u=2^x$

Fernando Revilla

3. Hello, paulaa!

Another approach . . .

$\displaystyle 2^{3-x} \:=\: 2^x - 2$

Multiply through by $\displaystyle 2^x:\;\;2^3 \:=\:2^{2x} - 2\!\cdot\!2^x \quad\Rightarrow\quad 2^{2x} - 2\!\cdot\!2^x - 8 \:=\:0$

Factor: .$\displaystyle (2^x + 2)(2^x-4) \:=\:0$

and we have: .$\displaystyle \begin{Bmatrix}2^x+2 \:=\:0 & \Rightarrow & 2^x \,=\,\text{-}2 & \Rightarrow & \text{no real roots} \\ \\[-3mm] 2^x - 4 \:=\:0 & \Rightarrow & 2^x \,=\,4 & \Rightarrow & \boxed{x \,=\,2} \end{Bmatrix}$

4. It's worth noting that Fernando and Soroban's methods are essentially the same. Soroban just didn't formally make a substutution. Most students seem to understand the formal substitution better at first, and then switch to what Soroban did after they've done enough problems.