## anonymous one year ago solve for x: log(5)(11-6x) = log(5)(1-x)

1. Nnesha

hint: $\huge\rm log_b x = \log_b y$ $\huge\rm\cancel { log_b} x = \cancel{\log_b} y$ x=y

2. anonymous

Ohhh... X = 2 right?

3. anonymous

Or is it no solution?

Is 5 it's base

5. Nnesha

sorry i forgot :(

6. anonymous

Yes 5 is the base

7. anonymous

That's ok Nnesha. Thanks for trying to help :)

8. Nnesha

no i'm saying i forgot i ws helping u :P

9. anonymous

Oh hahaha

10. Nnesha

alright so what was your first step ?? :-)

11. anonymous

I would cross out the bases because they are the same right?

12. Nnesha

yep right and then simple algebra!!! :-)

16-6x=1-x 15=5x x=?

14. Nnesha

and yes you got it right :-)

15. mathmate

Hint: dom log(x) = (0,inf)

16. anonymous

Oh this is a tricky no solution problem isn't it....

17. mathmate

... because?

18. anonymous

The base is 5 instead of 10?

19. mathmate

not exactly. Use the hint!

20. anonymous

Oh! I would have no 'y' value right?

21. Nnesha

nope when there is just log then u have to suppose 10 base

22. mathmate

actually, the hint applies to any base, so it was not indicated.

23. anonymous

Oh ok, so the answer would just be x = 2?

24. mathmate

25. anonymous

26. Nnesha

i'm confused do you have to solve for x right ?

27. mathmate

Hint: Whenever you solve an equation involving log, you need to substitute back into the equation to reject all roots that make log negative (recall dom log(x)=(0,inf) ).

28. mathmate

@Nnesha yes, the solution x=2 is correct for the algebraic part, but not correct for the problem given!

29. anonymous

I have to solve for the problem given @Nnesha

30. anonymous

Oh ok, so when I plug x=2 into the final product, I get f(2) = (0,inf) which is not solvable so there is no solution

31. mathmate

$$log_5(11-6x)=log_5(-1)$$ so x=2 is not an admissible solution because log(-1) is outside the domain of log.

32. mathmate

* -1 is outside the domain of log.

33. mathmate

sorry, gtg. be back later if you still have questions.

34. anonymous

I understand now! Thank you for bearing with me (haha) and helping me through all the steps!!!

35. mathmate

You're welcome! :)