# Unknown in Indices

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• Jan 27th 2008, 05:28 PM
Ogsta101
Unknown in Indices
I am having so much trouble with these kinds of equations. Can someone please help me:confused:

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

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

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
$

Sorry about all the confusion...

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
$

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
• 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
$

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:

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

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