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youngmath

  • 2 years ago

solve without using calculator 5ln(x-2)-2=28 help please, thank you

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  1. Champs
    • 2 years ago
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    5ln(x-2) = 30

  2. Champs
    • 2 years ago
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    ln(x-2) = 6

  3. Champs
    • 2 years ago
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    ln or log ?

  4. Mimi_x3
    • 2 years ago
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    \( 5ln(x-2) = 30 \) \(ln(x-2) = 6\) \(log_e(x-2) = 6\) \((x-2) = e^{6}\) Are you able to do it now?

  5. youngmath
    • 2 years ago
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    ln

  6. youngmath
    • 2 years ago
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    im stuck at the (x-2)=e^6

  7. Mimi_x3
    • 2 years ago
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    bring the \(2\) to the RHS

  8. youngmath
    • 2 years ago
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    i know the answer is e^6 +2 but i dont know where the =2 comes from

  9. youngmath
    • 2 years ago
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    +2

  10. Mimi_x3
    • 2 years ago
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    Because when you bring the \(-2\) to the RHS it's a positive

  11. Mimi_x3
    • 2 years ago
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    I don't know the correct way to say that lol

  12. youngmath
    • 2 years ago
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    alright can you walk me through the part after ln(x-2)=6

  13. Mimi_x3
    • 2 years ago
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    Well you know that \(ln\) can also be written as \(log_e\) right?

  14. youngmath
    • 2 years ago
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    yea

  15. youngmath
    • 2 years ago
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    so you get e^6= (x-2) right

  16. Mimi_x3
    • 2 years ago
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    yeah \((x-2) = e^{6}\)

  17. youngmath
    • 2 years ago
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    so what do you do next

  18. Mimi_x3
    • 2 years ago
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    well i just told you before :)

  19. youngmath
    • 2 years ago
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    bring the 2 to the RHS?

  20. youngmath
    • 2 years ago
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    ooooohhhhhhhh xD duh x = e^6 + 2

  21. Mimi_x3
    • 2 years ago
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    yes; you dont really bring it was called something else which forgot. just simply drag it to RHS. But there is a better name for it lol

  22. youngmath
    • 2 years ago
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    i forgot i was solving for x , silly me

  23. youngmath
    • 2 years ago
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    i more last question what do you do for e^x-3 = 16

  24. Mimi_x3
    • 2 years ago
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    \[ e^x - 3 =16\] What have you tried?

  25. youngmath
    • 2 years ago
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    e^x-3 + 4 =20

  26. Mimi_x3
    • 2 years ago
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    Um, that is not right..

  27. Mimi_x3
    • 2 years ago
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    drag the \(-3) to the rhs

  28. youngmath
    • 2 years ago
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    but its an exponent x-3 is the exp. for e

  29. Mimi_x3
    • 2 years ago
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    \[ e^x = 16+ 3\]

  30. Mimi_x3
    • 2 years ago
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    \[ e^x = 19\]

  31. Mimi_x3
    • 2 years ago
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    Take ln both sides \[xlne = ln19\]

  32. youngmath
    • 2 years ago
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    the answer is ln(16)+3

  33. Mimi_x3
    • 2 years ago
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    Wolfram Alpha agrees with me http://www.wolframalpha.com/input/?i=+e%5Ex-3+%3D+16

  34. youngmath
    • 2 years ago
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    my teacger myst be wring though . she gave me that answer

  35. youngmath
    • 2 years ago
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    my teacher was wron. she gave me that answer

  36. Mimi_x3
    • 2 years ago
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    well there is a low chance that wolfram alpha is wrong

  37. m_charron2
    • 2 years ago
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    oh, well, it looks like we have a paranthesis confusion here. when writing e^x-3 = 16, it's important that you write the parantheses aroudnthe (x-3), else we can't know that the whole thing is the exponant. So we have this right? \[e ^{x-3} = 16\]

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