# Unknown in Indices

• Jan 27th 2008, 05:28 PM
Ogsta101
Unknown in Indices

Solve for 'x':

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

and...

2^2x - 5.2^x+1 + 16 = 0
• Jan 27th 2008, 05:38 PM
Jhevon
Quote:

Originally Posted by Ogsta101

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

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

?
• Jan 27th 2008, 05:43 PM
mr fantastic
Quote:

Originally Posted by Ogsta101

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 = .......
• Jan 27th 2008, 05:48 PM
Ogsta101
For the first, I mean:

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

And thanx for help on number 2!
• Jan 27th 2008, 05:50 PM
Jhevon
Quote:

Originally Posted by Ogsta101
For the first, I mean:

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

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
• Jan 27th 2008, 05:56 PM
mr fantastic
Quote:

Originally Posted by Ogsta101
For the first, I mean:

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

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:

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

$\Rightarrow (2^{x})^2 - 6 \cdot 2^{x} + 8 = 0$
• Jan 27th 2008, 06:00 PM
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
• Jan 27th 2008, 06:09 PM
mr fantastic
Quote:

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

(Yes)