4(2^2x) + 31 (2^x) - 8 = 0.... please help.

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(2^{2x}) + 31 (2^{x}) - 8 = 0

this is what I have done so far...

4(2^{2x}) + 31 (2^{x}) - 8 = 0

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

4z^{2} +31z- 8 = 0 ( by replacing 2^{x} 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(3^{2x}) - 28(3^{x}) + 9 = 0

3 ((3^{x})^{2}) - 28 (3^{x}) + 9 = 0

3z^{2} - 28z + 9 = 0

3z^{2}- 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

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

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!

THANKS IN ADVANCED!

Re: 4(2^2x) + 31 (2^x) - 8 = 0.... please help.

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:

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

So,

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!

Re: 4(2^2x) + 31 (2^x) - 8 = 0.... please help.

Quote:

Originally Posted by

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

I have to solve 4(2^{2x}) + 31 (2^{x}) - 8 = 0

this is what I have done so far...

4(2^{2x}) + 31 (2^{x}) - 8 = 0

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

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

so