Log question

**Krislton** · March 9th 2013, 10:39 AM

Please can someone help with this problem: solve $2^2^x^+^1 - 5 (2^x) + 2 = 0$ 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

**Shakarri** · March 9th 2013, 10:55 AM

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)²Edit: Beat plato by 1 minute :P

**Plato** · March 9th 2013, 10:56 AM

Originally Posted by Krislton

Please can someone help with this problem: solve 2^(2x+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?

If we let $y=2^x$ that equation becomes $2y^2-5y+2=0$ .

**Krislton** · March 9th 2013, 11:02 AM

Originally Posted by Shakarri

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

Hint: 2^2x+1=2(2^x)²

Yeah I think your correct, but I'm still useless at logs! Its not from a text book, but from an old exam paper I need to get to grips with ASAP. Once re-arranged how you've done it, Should I then apply logs to both sides?

Thanks for your help,

Kris.

**Krislton** · March 9th 2013, 11:15 AM

Originally Posted by Plato

If we let $y=2^x$ that equation becomes $2y^2-5y+2=0$ .

Thanks, it kind of makes sense but if substituting 2^x = y, how can you sub that for 2x when that's not to the power unlike the other term.

**Soroban** · March 9th 2013, 01:17 PM

Hello, Krislton!

Evidently you don't understand the hint or the suggestions.
I'll give you a walk-through . . .

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

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

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

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

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

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

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

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

**Krislton** · March 9th 2013, 01:53 PM

Originally Posted by Soroban

Hello, Krislton!

Evidently you don't understand the hint or the suggestions.
I'll give you a walk-through . . .

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

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

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

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

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

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

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

Perfect, thanks very much I understand now!

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