Hi, i need some help solving the following equation, for some reason i can't get my head around it:
4(2^2x)+31(2^x)-8=0
it would really help me if you can post an explantion on how you solved it from the begining to the end.
Hi, i need some help solving the following equation, for some reason i can't get my head around it:
4(2^2x)+31(2^x)-8=0
it would really help me if you can post an explantion on how you solved it from the begining to the end.
Hello, wolflessalex!
$\displaystyle 4\!\cdot\!2^{2x}+31\!\cdot\!2^x-8\:=\:0$
We have a quadratic: .$\displaystyle 4(2^x)^2 + 31(2^x) - 8 \;=\;0$
Let $\displaystyle u = 2^x$
Substitute: .$\displaystyle 4u^2 + 31u - 8 \;=\;0$
Factor: .$\displaystyle (u+8)(4u-1) \:=\:0$
And we have: .$\displaystyle \begin{Bmatrix}u+8\:=\:0 & \Rightarrow & u \:=\:\text{-}8 \\4u-1\:=\:0 & \Rightarrow & u \:=\:\frac{1}{4} \end{Bmatrix}$
Back-substitute:
. . $\displaystyle 2^x \:=\:\text{-}8 \quad \text{No real root}$
. . $\displaystyle 2^x \:=\:\tfrac{1}{4} \quad\Rightarrow\quad & 2^x \:=\:2^{-2} \quad\Rightarrow\quad x \:=\:\text{-}2 $