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 ?
$\displaystyle 2^{3-x}=2^x-2\;\Leftrightarrow\;\dfrac{2^3}{2^x}=2^x-2$
Now, use the substitution $\displaystyle u=2^x$
Fernando Revilla
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: .$\displaystyle \begin{Bmatrix}2^x+2 \:=\:0 & \Rightarrow & 2^x \,=\,\text{-}2 & \Rightarrow & \text{no real roots} \\ \\[-3mm]
2^x - 4 \:=\:0 & \Rightarrow & 2^x \,=\,4 & \Rightarrow & \boxed{x \,=\,2} \end{Bmatrix}$
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.