1. ## Unknown in Indices

Solve for 'x':

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

and...

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

2. 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

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

?

3. 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 = .......

4. For the first, I mean:

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

And thanx for help on number 2!

5. 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

6. 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:

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

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

7. 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

8. 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