Unknown in Indices

**Ogsta101** · January 27th 2008, 04:28 PM

I am having so much trouble with these kinds of equations. Can someone please help me

Solve for 'x':

2^2x-1 - 6.2^x-1 + 4 = 0

and...

2^2x - 5.2^x+1 + 16 = 0

**Jhevon** · January 27th 2008, 04:38 PM

Originally Posted by Ogsta101

I am having so much trouble with these kinds of equations. Can someone please help me

Solve for 'x':

2^2x-1 - 6.2^x-1 + 4 = 0

and...

2^2x - 5.2^x+1 + 16 = 0

please clarify (use parentheses!) i'm not sure what the questions are...

is it:

$2^{2x - 1} - 6 \cdot 2^{x - 1} + 4 = 0$

and

$2^{2x} - 5 \cdot 2^{x + 1} + 16 = 0$

?

**mr fantastic** · January 27th 2008, 04:43 PM

Originally Posted by Ogsta101

I am having so much trouble with these kinds of equations. Can someone please help me

Solve for 'x':

2^2x-1 - 6.2^x-1 + 4 = 0

and...

2^2x - 5.2^x+1 + 16 = 0

For the first, do you mean $2^{2x - 1} - 6 \times 2^x + 4 = 0 \,$ ? Otherwise you have $2^{2x - 1}$ appearing twice .......

For the second:

$2^{2x} - 5 \times 2^{x+1} + 16 = 0$

$\Rightarrow (2^{x})^2 - 5 \times 2 \times 2^{x} + 16 = 0$

$\Rightarrow (2^{x})^2 - 10 \times 2^{x} + 16 = 0$

Let $w = 2^x$ :

$\Rightarrow w^2 - 10w + 16 = 0$

$\Rightarrow (w - 2)(w - 8) = 0$

$\Rightarrow w = 2, \, 8$

Therefore $2 = 2^x \,$ or $8 = 2^x$ .

Therefore $2^1 = 2^x \,$ or $2^3 = 2^x$ .

Therefore x = .......

**Ogsta101** · January 27th 2008, 04:48 PM

For the first, I mean:

$ 2^{2x - 1} - 6 \cdot 2^{x - 1} + 4 = 0 $

Sorry about all the confusion...

And thanx for help on number 2!

**Jhevon** · January 27th 2008, 04:50 PM

Originally Posted by Ogsta101

For the first, I mean:

$ 2^{2x - 1} - 6 \cdot 2^{x - 1} + 4 = 0 $

Sorry about all the confusion...

And thanx for help on number 2!

well, it's very similar to what mr fantastic did. same concept. try and see if you can do it

**mr fantastic** · January 27th 2008, 04:56 PM

Originally Posted by Ogsta101

For the first, I mean:

$ 2^{2x - 1} - 6 \cdot 2^{x - 1} + 4 = 0 $

Sorry about all the confusion...

And thanx for help on number 2!

Just to get you started:

$2^{2x - 1} - 6 \cdot 2^{x - 1} + 4 = 0$

Trick - Multiply through by 2 to make life easier:

$\Rightarrow 2^{2x} - 6 \cdot 2^{x} + 8 = 0$

$\Rightarrow (2^{x})^2 - 6 \cdot 2^{x} + 8 = 0$

**Ogsta101** · January 27th 2008, 05:00 PM

Well, I did it without the multiplying through by 2 and got
x = 2 or 1
and they appear to work.

Thanx heaps for helping me out

**mr fantastic** · January 27th 2008, 05:09 PM

Originally Posted by Ogsta101

Well, I did it without the multiplying through by 2 and got
x = 2 or 1
and they appear to work.

Thanx heaps for helping me out

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