Thread: 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.)

4z² +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

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

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!

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!

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.)

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

so