## StudyGurl14 one year ago HELP PLEASE Solving for x.

1. StudyGurl14

|dw:1440620935256:dw| Obviously, I did something wrong.

2. StudyGurl14

@perl

3. perl

|dw:1440621059469:dw|

4. StudyGurl14

Oh, I see. I think.

5. perl

multiply both sides by 2^x, to get rid of 2^x in the denominator

6. StudyGurl14

|dw:1440621096683:dw| Like that?

7. perl

|dw:1440621179162:dw|

8. freckles

hint: you will wind up with a quadratic in terms of 2^x

9. perl

|dw:1440621199302:dw|

10. StudyGurl14

Oh, I forgot the 1.

11. StudyGurl14

so...? |dw:1440621240821:dw|

12. StudyGurl14

???

13. StudyGurl14

Now do I use ln?

14. perl

|dw:1440621466522:dw|

15. StudyGurl14

Oh, so 2^2x

16. perl

|dw:1440621493371:dw|

17. StudyGurl14

|dw:1440621524603:dw| ???

18. perl

Now you have a quadratic equation.

19. StudyGurl14

How?^

20. phi

FYI, what you did wrong is that when you start with $2^x + 2^{-x}= 5$ we take the ln of both sides $\ln \left( 2^x + 2^{-x}\right) = \ln 5$ and that is *not* $\ln \left( 2^x\right) + \ln \left( 2^{-x}\right) = \ln 5$

21. StudyGurl14

Okay, so how do I distrubite the ln @phi ?

22. perl

if we let u = 2^x. then we have u^2 - 5*u +1 = 0

23. perl

you can't distribute the ln

24. StudyGurl14

Oh, I see @perl

25. StudyGurl14

so.... 5/2 +- sqrt{21}/2 ?

26. phi

yes, and now you can do 2^x = mess and take the log of both sides

27. StudyGurl14

what's "mess" ?

28. phi

5/2 +- sqrt{21}/2

29. StudyGurl14

okay, so |dw:1440621917929:dw| like that?

30. perl

we want to solve (2^x)^2 - 5*(2^x) +1 = 0 if we let u = 2^x. then we have u^2 - 5*u +1 = 0 the solutions are u = 5/2 + sqrt{21}/2 , 5/2 - sqrt{21}/2 Now back substitute since we really wanted to solve for x 2^x = 5/2 + sqrt{21}/2 2^x = 5/2 - sqrt{21}/2 now take ln of both sides ln (2^x)= ln( 5/2 + sqrt{21}/2)

31. StudyGurl14

I got approx 2.26?

32. perl

yes

33. StudyGurl14

Okay, thank you.

34. perl

and the other solution?

35. StudyGurl14

Do I do it again for the negative one?

36. perl

right

37. StudyGurl14

so approx -2.26

38. perl

indeed :)

39. StudyGurl14

Thank you!

40. perl

and as phi pointed out, in general ln( x + y) does not equal ln(x) + ln (y) same with other many other functions sin ( x + y) does not equal sin(x) + sin(y),