Find the solution set of 2^(2x-2)-2*2^(x-1)=8

[Jaco Viljoen]

Homework Statement

Find the solution set of 2^(2x-2)-2*2^(x-1)=8

Homework Equations

The Attempt at a Solution

2^(2x-2)-2*2^(x-1)=8
2^(2x-2)-2^(x-1+1)=8
2^(2x-2)-2^(x)=8
2^(2x-2)-2^(x)-8=0
2^(2(x-1))-2^(x)-8=0
I cannot solve the equation,

I just need direction with this step and will attempt the rest on my own, thank you.
Jaco

 

[BvU]

Re grasping: you can do this: 2*2^(x-1) = 2^x.
And you can do something with 2^2x too ! Once you see that, it's easy ...

 

[Jaco Viljoen]

Re grasping: you can do this: 2*2^(x-1) = 2^x.
And you can do something with 2^2x too ! Once you see that, it's easy ...

Thank you BvU

2^(2x-2)-2*2^(x-1)=8
2^(2x-2)-2^(x-1+1)=8
2^(2x-2)-2^(x)=8
2^(2x-2)-2^(x)-8=0
2^(2(x-1))-2^(x)-8=0
4^(x-1)-2^x=8

or
2^(2x-2)-2*2^(x-1)=8
2^(2(x-1))-2*2^(x-1)=8
4^(x-1)-2*2^(x-1)=8
4^(x-1)-(2*2^(x-1))=8
4^(x-1)-2^(x-1+1)=8
2^(2(x-1))-2^x=8
(2^x-1)²-2^x=8
ax²+bx+c=0
(2^x-1)²-2^x-8=0

 

[certainly]

Hi Jaco,
You should use either the sub-script and super-script buttons from the post template or ##LATEX##, it will make your posts a lot more readable.
The question is basically asking which two powers of 2 have the difference 8. If you just want an answer look at the differences of the first few powers of 2.
To prove it algebraically, you could either factor the expression ##2^{2x-2}-2^x## or continuing on what BvU said, modify the expression ##4^{x-1}## a little more and it should become familiar territory.

 

[HallsofIvy]

The first thing I would do is let [itex]y= 2^{x- 1}[/itex].

[mentor note: content abridged]

... see how you go with that substitution.

 

[BvU]

Yes 2^2x = (2²)^x. But it's also 2^x2 = ... Aha !

 

[Jaco Viljoen]

The first thing I would do is let [itex]y= 2^{x- 1}[/itex]. [mentor note: content deleted]

Hi Hallsoflvy,
how do you get the 2y? b=-2^x not 2^x-1
or should I be stepping back to:
2^(x-1)2-2(2^x-1)=8
ax²+bx+c=0
y²-2y-8=0
y²-2y-8=0

 

[certainly]

Hi Hallsoflvy,
how do you get the 2y? b=-2^x not 2^x-1
or should I be stepping back to:
2^(x-1)2-2(2^x-1)=8

HallsofIvy took ##b=2^{x-1}## BvU and I took ##b=2^{x}##.

 

[Jaco Viljoen]

HallsofIvy took ##b=2^{x-1}## BvU and I took ##b=2^{x}##.

I also used b=2^x
But I will attempt this way.

 

[Jaco Viljoen]

Now note that if x>3x>3 then 2x−2−12^{x-2}-1 will be odd, however 8 cannot have an odd factor.
But if x<3x

Hi Certainly,
where did you get 3?
I know you see it must be 3, but how is it shown?
Should I not have proved this?

 

[certainly]

In order for ##2^{x-2}-1>0## the power of 2 must be at-least 1. Therefore ##x-2=1## or ##x=3##.

 

[Jaco Viljoen]

2^2x-2-2*2^x-1=8
2^2x-2-2^x-1+1=8
2^2x-2-2^x=8
2^x(2^x-2-1)=8
for 2^x(2^x-2-1)>0 the power of 2<=1
x=3
{3}
Is that how I would end my calculation?

 

[SammyS]

Hi Hallsoflvy,
how do you get the 2y? b=-2^x not 2^x-1
or should I be stepping back to:
2^(x-1)2-2(2^x-1)=8
ax²+bx+c=0
y²-2(2y)-8=0 This is incorrect.
y²-4y-8=0

2^(x-1)2 is equivalent to (2^(x-1))² . You did get this right, but the extra parentheses makes clear what is intended.

However, if y = 2^x-1 , then 2(2^x-1) = 2(y), not 4(y) .

 

[Jaco Viljoen]

Thank you Sammy,