Math Help - Log question

1. Log question

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?

Kris

2. Re: Log question

You might have misread your book and it is telling you to make the substitution y=2x
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: 22x+1=2(2x)2

​Edit: Beat plato by 1 minute :P

3. Re: Log question

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 $2y^2-5y+2=0$.

4. Re: Log question

Originally Posted by Shakarri
You might have misread your book and it is telling you to make the substitution y=2x
If not, do that substitution anyway and solve for y

Hint: 22x+1=2(2x)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?

Kris.

5. Re: Log question

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.

6. Re: Log question

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

7. Re: Log question

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!