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

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- Jan 27th 2008, 04:28 PMOgsta101Unknown 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, 04:38 PMJhevon
- Jan 27th 2008, 04:43 PMmr fantastic
For the first, do you mean $\displaystyle 2^{2x - 1} - 6 \times 2^x + 4 = 0 \,$ ? Otherwise you have $\displaystyle 2^{2x - 1}$ appearing twice .......

For the second:

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

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

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

Let $\displaystyle w = 2^x$:

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

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

$\displaystyle \Rightarrow w = 2, \, 8$

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

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

Therefore x = ....... - Jan 27th 2008, 04:48 PMOgsta101
For the first, I mean:

$\displaystyle

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, 04:50 PMJhevon
- Jan 27th 2008, 04:56 PMmr fantastic
- Jan 27th 2008, 05:00 PMOgsta101
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, 05:09 PMmr fantastic