## youngmath 3 years ago solve without using calculator 5ln(x-2)-2=28 help please, thank you

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1. Champs

5ln(x-2) = 30

2. Champs

ln(x-2) = 6

3. Champs

ln or log ?

4. Mimi_x3

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

5. youngmath

ln

6. youngmath

im stuck at the (x-2)=e^6

7. Mimi_x3

bring the \(2\) to the RHS

8. youngmath

i know the answer is e^6 +2 but i dont know where the =2 comes from

9. youngmath

+2

10. Mimi_x3

Because when you bring the \(-2\) to the RHS it's a positive

11. Mimi_x3

I don't know the correct way to say that lol

12. youngmath

alright can you walk me through the part after ln(x-2)=6

13. Mimi_x3

Well you know that \(ln\) can also be written as \(log_e\) right?

14. youngmath

yea

15. youngmath

so you get e^6= (x-2) right

16. Mimi_x3

yeah \((x-2) = e^{6}\)

17. youngmath

so what do you do next

18. Mimi_x3

well i just told you before :)

19. youngmath

bring the 2 to the RHS?

20. youngmath

ooooohhhhhhhh xD duh x = e^6 + 2

21. Mimi_x3

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

i forgot i was solving for x , silly me

23. youngmath

i more last question what do you do for e^x-3 = 16

24. Mimi_x3

\[ e^x - 3 =16\] What have you tried?

25. youngmath

e^x-3 + 4 =20

26. Mimi_x3

Um, that is not right..

27. Mimi_x3

drag the \(-3) to the rhs

28. youngmath

but its an exponent x-3 is the exp. for e

29. Mimi_x3

\[ e^x = 16+ 3\]

30. Mimi_x3

\[ e^x = 19\]

31. Mimi_x3

Take ln both sides \[xlne = ln19\]

32. youngmath

33. Mimi_x3

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

34. youngmath

my teacger myst be wring though . she gave me that answer

35. youngmath

my teacher was wron. she gave me that answer

36. Mimi_x3

well there is a low chance that wolfram alpha is wrong

37. m_charron2

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? \[e ^{x-3} = 16\]