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anonymous

  • one year ago

solve for x: log(5)(11-6x) = log(5)(1-x)

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  1. Nnesha
    • one year ago
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    hint: \[\huge\rm log_b x = \log_b y\] \[\huge\rm\cancel { log_b} x = \cancel{\log_b} y\] x=y

  2. anonymous
    • one year ago
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    Ohhh... X = 2 right?

  3. anonymous
    • one year ago
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    Or is it no solution?

  4. Pawanyadav
    • one year ago
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    Is 5 it's base

  5. Nnesha
    • one year ago
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    sorry i forgot :(

  6. anonymous
    • one year ago
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    Yes 5 is the base

  7. anonymous
    • one year ago
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    That's ok Nnesha. Thanks for trying to help :)

  8. Nnesha
    • one year ago
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    no i'm saying i forgot i ws helping u :P

  9. anonymous
    • one year ago
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    Oh hahaha

  10. Nnesha
    • one year ago
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    alright so what was your first step ?? :-)

  11. anonymous
    • one year ago
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    I would cross out the bases because they are the same right?

  12. Nnesha
    • one year ago
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    yep right and then simple algebra!!! :-)

  13. Pawanyadav
    • one year ago
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    16-6x=1-x 15=5x x=?

  14. Nnesha
    • one year ago
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    and yes you got it right :-)

  15. mathmate
    • one year ago
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    Hint: dom log(x) = (0,inf)

  16. anonymous
    • one year ago
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    Oh this is a tricky no solution problem isn't it....

  17. mathmate
    • one year ago
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    ... because?

  18. anonymous
    • one year ago
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    The base is 5 instead of 10?

  19. mathmate
    • one year ago
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    not exactly. Use the hint!

  20. anonymous
    • one year ago
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    Oh! I would have no 'y' value right?

  21. Nnesha
    • one year ago
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    nope when there is just log then u have to suppose 10 base

  22. mathmate
    • one year ago
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    actually, the hint applies to any base, so it was not indicated.

  23. anonymous
    • one year ago
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    Oh ok, so the answer would just be x = 2?

  24. mathmate
    • one year ago
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    want further hint? @lehmad

  25. anonymous
    • one year ago
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    yes please :p

  26. Nnesha
    • one year ago
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    i'm confused do you have to solve for x right ?

  27. mathmate
    • one year ago
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    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
    • one year ago
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    @Nnesha yes, the solution x=2 is correct for the algebraic part, but not correct for the problem given!

  29. anonymous
    • one year ago
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    I have to solve for the problem given @Nnesha

  30. anonymous
    • one year ago
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    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
    • one year ago
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    \(log_5(11-6x)=log_5(-1)\) so x=2 is not an admissible solution because log(-1) is outside the domain of log.

  32. mathmate
    • one year ago
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    * -1 is outside the domain of log.

  33. mathmate
    • one year ago
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    sorry, gtg. be back later if you still have questions.

  34. anonymous
    • one year ago
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    I understand now! Thank you for bearing with me (haha) and helping me through all the steps!!!

  35. mathmate
    • one year ago
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    You're welcome! :)

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