## open_study1 3 years ago If x= z and y=a satisfys the equation 2^(x+y) = 6^y and 3^(x-1)= 2^(y+1) then z+a=?

1. open_study1

options are \[\log_{3}6 \] \[\log_{3/2} 6\]

2. open_study1

3. shivam_bhalla

Take log on both sides \[\large \log(2^{x+y}) = \log(6^y)\] Applying power rule, we get (x+y)log 2 = y log 6 Similarly apply for the 2nd equation and get two equations and try to solve

4. experimentX

\[ 6 = 2^{log_2 6} and \; 3 = 2^{log_2 3}\] do some algebric manipulation

5. shivam_bhalla

Also log 6 = log(2*3) = log 2 + log 3

6. open_study1

i am stuck @experimentX @satellite73 @Callisto plz help

7. experimentX

just some algebraic manipulation ... cancel out the bases .... equate the powers two eqautions ... two unkonwsn ... solve them!!

8. open_study1

oh......no !!!!! getting an error plzz help

9. siddhantsharan

Yeah you should get : \[\log_{3/2}6 \]

10. open_study1

the correct answer is that which @siddhantsharan wrote

11. experimentX

\[ (x+y) = y\log_2 6 \] \[ (x-1)\log_23 = (y+1)\] try solving them

12. siddhantsharan

Dont solve them. Manupilate to get x+ y.

13. siddhantsharan

Take everything on a base 3 by taking log on both sides. \[(x + y)\log_{3}2 = y \log_{3} 6\] And x -1 = log 2(y +1)

14. siddhantsharan

From the first you get x log2 = y

15. siddhantsharan

Then Add the equations And solve for x + y

16. open_study1
17. open_study1

@shivam_bhalla u visited my question.

18. open_study1

@satellite73 @Mertsj

19. Mertsj

I added the equations like you said but couldn't make any progress from there.

20. siddhantsharan

Okay. Gimme a minute to type it. :)

21. siddhantsharan

Assume all logs as base 3 unless written otherwise. \[y = xlog2 \] \[x - 1 = ylog2 - \log2\] Adding, \[(x + y) - 1 = (\log2)(x+y)\]

22. siddhantsharan

Let x + y be t. t - 1 = tlog2 => t = 1 / ( 1 - log2)

23. siddhantsharan

|dw:1338210511343:dw|

24. siddhantsharan

@Mertsj Did you understand?

25. Mertsj

I think so except I thought the one equation was \[x-1=\log2(y+1)\]

26. siddhantsharan

Oops. My bad. It is y + 1. Sorry. Steps remain same except answer becomes log(base 3/2) 6

27. Mertsj

I think that now I can figure it out if I study it. Thank you so much. I have been working on this for a long time. Don't ask me why because I am just an old lady and it makes no difference at all. Thank so much.

28. siddhantsharan