• Feb 4th 2013, 06:23 PM
marrzbarrz
Hi, I am not sure if I am in the correct place but here I go....

I am taking Advance Functions, and I am currently in a unit about Exponential and Logarithmic Equations.

I have to solve 4(22x) + 31 (2x) - 8 = 0

this is what I have done so far...

4(22x) + 31 (2x) - 8 = 0
4((2x)2) + 31 (2x) - 8 = 0 (the book says " you can do a substitution here to make things a little easier.)

4z2 +31z- 8 = 0 ( by replacing 2x with Z) ***I am stuck here***

Here is a similar example the book has... (i used it to help guide me through this one, but i couldn't get pass the 31z.)

3(32x) - 28(3x) + 9 = 0

3 ((3x)2) - 28 (3x) + 9 = 0

3z2 - 28z + 9 = 0

3z2- 27z - z + 9 = 0

3z(z-9) - (z-9) = 0

(3z -1) (z-9) = 0 (by factoring)

3z -1 = 0 or z - 9 = 0

z= 1/3 or z= 9

3x = 1/3 or 3x= 9 (by replacing z with 3x)

Therefore, x = -1 or x = 2.

Anyways I really do hope someone can help with this... I've been stuck on this problem for a couple days now & i just want to be finish!

• Feb 4th 2013, 06:44 PM
sakonpure6
For the first problem, I see a quadratic problem that you can solve either by factoring or using the quadratic formula!
Ill give this a go:

$4z^2 +31z- 8 = 0$

this can be factored. I know right away because I look at the discriminant which when solved yields a perfect square of 1089.

So,

$4z^2 +31z- 8 = 0$
$(4z -1)(1z +8 )=0$

There for z = 1/4 and -8

I only took grade 11 functions, but just looking at your problem, I hope what I did helps!
• Feb 4th 2013, 06:46 PM
Plato
$4z^2+31z-8=(4z-1)(z+8)=0$ so $z=~?~\&~z=~?$