Solve the following exponential equation

**wolflessalex** · October 25th 2012, 02:15 PM

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.

**Soroban** · October 25th 2012, 02:52 PM

Hello, wolflessalex!

$4\!\cdot\!2^{2x}+31\!\cdot\!2^x-8\:=\:0$

We have a quadratic: . $4(2^x)^2 + 31(2^x) - 8 \;=\;0$

Let $u = 2^x$

Substitute: . $4u^2 + 31u - 8 \;=\;0$

Factor: . $(u+8)(4u-1) \:=\:0$

And we have: . $\begin{Bmatrix}u+8\:=\:0 & \Rightarrow & u \:=\:\text{-}8 \\4u-1\:=\:0 & \Rightarrow & u \:=\:\frac{1}{4} \end{Bmatrix}$

Back-substitute:

. . $2^x \:=\:\text{-}8 \quad \text{No real root}$

. . $2^x \:=\:\tfrac{1}{4} \quad\Rightarrow\quad & 2^x \:=\:2^{-2} \quad\Rightarrow\quad x \:=\:\text{-}2$

**wolflessalex** · October 25th 2012, 04:14 PM

Thank you very much!!! Your explanation was perfect!!
Finally i understood this annoying equation!!

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