Please can someone help with this problem: solve $\displaystyle 2^2^x^+^1 - 5 (2^x) + 2 = 0 $ Now we were told to let y=2 but I can't figure out where the y is coming from? Is it just as a function of x?

Thanks in advance for your help!

Kris :)

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- Mar 9th 2013, 10:39 AMKrisltonLog question
Please can someone help with this problem: solve $\displaystyle 2^2^x^+^1 - 5 (2^x) + 2 = 0 $ Now we were told to let y=2 but I can't figure out where the y is coming from? Is it just as a function of x?

Thanks in advance for your help!

Kris :) - Mar 9th 2013, 10:55 AMShakarriRe: Log question
You might have misread your book and it is telling you to make the substitution y=2

^{x}

If not, do that substitution anyway and solve for y, after you have solved for y then finish by using it to find x

Hint: 2^{2x+1}=2(2^{x})^{2 }Edit: Beat plato by 1 minute :P - Mar 9th 2013, 10:56 AMPlatoRe: Log question
- Mar 9th 2013, 11:02 AMKrisltonRe: Log question
- Mar 9th 2013, 11:15 AMKrisltonRe: Log question
- Mar 9th 2013, 01:17 PMSorobanRe: Log question
Hello, Krislton!

Evidently you don't understand the hint or the suggestions.

I'll give you a walk-through . . .

Quote:

$\displaystyle \text{Solve: }\:2^{2x+1} - 5 (2^x) + 2 \:=\: 0$

We have: .$\displaystyle 2^{2x+1} - 5(2^x) + 2 \;=\;0$

. . . . . . . . $\displaystyle 2^{2x}\!\cdot\! 2^1 - 5(2^x) + 2 \;=\;0$

. . . . . . . . $\displaystyle 2(2^x)^2 - 5(2^x) + 2 \;=\;0$

Let $\displaystyle y = 2^x\!:\qquad 2y^2 - 5y + 2 \;=\;0$

Factor: . . . . $\displaystyle (y-2)(2y-1) \;=\;0$

And we have: .$\displaystyle y = 2,\;y = \tfrac{1}{2}$

Back-substitute: .$\displaystyle \begin{Bmatrix}2^x \,=\,2 & \Rightarrow & 2^x \,=\,2^1 & \Rightarrow & x \,=\,1 \\ \\[-3mm] 2^x \,=\,\frac{1}{2} & \Rightarrow & 2^x \,=\,2^{\text{-}1} & \Rightarrow & x \,=\,\text{-}1 \end{Bmatrix}$

- Mar 9th 2013, 01:53 PMKrisltonRe: Log question