## anonymous 3 years ago 2log(x-3)+1 = 5 Could you break it down step by step? I'm a bit confused on how to do this algebraically

1. Callisto

To solve x?

2. anonymous

yes, i presume

3. Callisto

Okay, first, subtract both sides by 1, what do you get?

4. anonymous

you get 2log(x-3) = 4

5. Callisto

Right, now divide both sides by 2, what do you get?

6. anonymous

log(x-3) = 2

7. anonymous

haha, that's fine :)

8. calculusfunctions

Don't mind me. @Callisto seems to be doing a fine job teaching so I'll just observe if you don't mind? I'll only respond if you ask.

9. Callisto

Yup! Now, take anti-log for both sides. What do you get? Note: anti-log = inverse of log For example, logx = 1 => 10^(logx) = 10^1 => x = 10^1 = 10

10. anonymous

yikes, haha, i'm not for sure. I briefly remember anti-logs but I'm not for sure what do with them

11. Callisto

Do you understand that example?

12. anonymous

i'm starting to get it. logx = 1. I understand that but not after that

13. calculusfunctions

If I may, @Callisto perhaps he'd @Egirl01 would understand if you asked her to change the logarithm form to exponential form. @Egirl01 do you know how to do that.

14. anonymous

No, I'm sorry

15. anonymous

give me one second

16. anonymous

$$\large b^x=y \rightarrow log_by=x$$

17. anonymous

log^2 = x-3?

18. Callisto

logx = 1 is actually just an example... ---------------------------------- Intuitively, log a number is the value of the power of the base Let say, I have a number 100, which can be rewritten as 10^2 Now, I take log (base 10) of that number (100), so I'll get 2, since 10 to the power of 2 give me 100. (Sorry if I make it even worse...)

19. anonymous

No, I understand the piece. so if its simply log, then its naturally log10, and the anti-log of log10 is 1/10 right?

20. Callisto

anti-log of log 10 is 10. $\log^{-1}(log10) = 10$ anti-log is the inverse of log.

21. anonymous

Alright. So then it'd be 10 (x-3) = 2?

22. Callisto

No.... How did you get that?

23. anonymous

the anti-log of log 10 is 10, so log(x-3) = 2 would turn to 10(x-3) = 2? Obviously that is wrong. I'm a bit lost but you can continue onto the next part of solving the equation. I'll eventually figure what I'm doing wrong

24. Callisto

$log(x-3) = 2$Take anti-log for both sides: $\log^{-1}(log(x-3)) = \log^{-1}(2)$ Perhaps you can simplify the left first.

25. anonymous

I still don't quite follow, but continue on. I'll understand eventually :)

26. Callisto

Nope.. It's your turn to work on it... Which part you don't understand ?

27. anonymous

how exactly I create the anti-log from log(x-3)

28. Callisto

Because you are doing something on both sides... You create anti-log to undo the work of log...

29. anonymous

So the anti-log of log10 is 10, right? and the anti-log of 2 is what?

30. Callisto

31. anonymous

okay

32. Callisto

What is log10?

33. anonymous

log10 =1

34. Callisto

What is log100?

35. anonymous

log100 = 2

36. Callisto

Can we tell me what number do I have to take log on in order to get 1?

37. anonymous

10?

38. Callisto

Yes. So, what number do I have to take log on in order to get 2?

39. anonymous

100

40. Callisto

Yes! How did you get this answer?

41. anonymous

because you inversed it?

42. Callisto

Yes! So, now, can you answer your question? what is $$\log^{-1}2$$?

43. anonymous

It is 100

44. Callisto

Yes!

45. anonymous

So 10(x-3) = 100?

46. Callisto

No. Another set question questions... What is log10 (again)?

47. anonymous

log 10 is 1

48. Callisto

You got it right for the right side, it's wrong for the left.

49. Callisto

What is $$\log^{-1}log(10)$$?

50. anonymous

10

51. Callisto

Yes. What is $$\log^{-1}(log100)$$?

52. anonymous

2?

53. anonymous

WAIT

54. anonymous

100?

55. Callisto

Yes!

56. Callisto

What is $$\log^{-1} (log1000)$$?

57. anonymous

1000

58. Callisto

So, we can see that $$\log^{-1}$$ of log a number is the number itself, agree?

59. anonymous

yes

60. Callisto

So, $$\log^{-1} (\log y) = y$$, agree?

61. anonymous

yes

62. Callisto

So, what is $$\log^{-1} (log(x-3))$$?

63. anonymous

x-3?

64. Callisto

Yes!

65. Callisto

Can you simplify the following now? $\log^{-1}(log(x-3)) = \log^{-1}(2)$

66. anonymous

x-3 = 100?

67. Callisto

Yes! Now can you solve it?

68. anonymous

yes

69. anonymous

x = 103?

70. Callisto

Nice :D and Yes! Do you understand how to solve this type of question now?

71. anonymous

yes! Thank you soooo much for your help! I really appreciate it!

72. Callisto

You're welcome :)