## chika_tika3007 2 years ago 2^2x -6.2^x +8 =0

1. ganeshie8

say k = \(\large 2^x\)

2. ganeshie8

\(\large 2^{2x} -6.2^x +8 =0\) \(\large (2^{x})^2 -6.2^x +8 =0\)

3. ganeshie8

plugin k and you will see a quadratic in k solve k

4. chika_tika3007

ok but how abaout the number 8???

5. ganeshie8

wat do u get after putting \(2^x=k\) ?

6. ganeshie8

we get this :- \(\large k^2-6k+8=0\)

7. ganeshie8

this is just a quadratic equation you knw how to solve a quadratic equaiton ha ?

8. chika_tika3007

yes

9. ganeshie8

good, factor it and solve k first

10. chika_tika3007

humm ok ok i got it k thanks

11. ganeshie8

sure ?

12. ganeshie8

\(\large (k-2)(k-4) = 0\) \(k = 2, k=4\)

13. chika_tika3007

i get the answer are -2 and -4

14. chika_tika3007

yes..

15. ganeshie8

when k = 2, \(2^x = k\) \(2^x = 2^1\) \(x = 1\)

16. ganeshie8

when k = 4, \(2^x = k\) \(2^x = 4\) \(2^x = 2^2\) \(x = 2\)

17. ganeshie8

so the solutions are : 1, 2

18. chika_tika3007

yup yup

19. ganeshie8

cool :)