## anonymous 4 years ago Can someone please explain how to solve a log equation with an example.

1. anonymous

how 'bout u pick an example....

2. anonymous

otherwise i'll choose the easiest...:)

3. anonymous

OK ... $5^{x - 3} . 3^{2x-8} = 225$

4. anonymous

How to solve this using log? I'm new so I have no idea.

5. anonymous

I mean new to log.

6. anonymous

pls help.

7. lgbasallote

welcome to logarithms!

8. anonymous

Yeah thanks!

9. anonymous

I guess you can solve this without using Logs..

10. anonymous

But I want to know how to do it with log.

11. anonymous

Okay then all here will explain you how to do this...

12. anonymous

15^(x-3)+(2x-8)=225 now we take ln (3x-11)15= ln225

13. anonymous

What is In? And how did you get 15^(x-3)+(2x-8)=225?

14. anonymous

15*ln(3x-11)=ln 225

15. anonymous

by multi 3 into 5 to get 15 and the exponent it's added

16. anonymous

But how can you just multiply the base and add the exponents?

17. anonymous

Use the following formulas of Logs: $\large Log(a \times b) = Log(a) + Log(b)$ $\large Log(a)^b = b.Log(a)$

18. anonymous

before we take the logarithem

19. anonymous

it's from exponent experepties

20. anonymous

Why did you delete that @dpaInc?

21. anonymous

because i don't wanna take away from these guys awesome explanations....:)

22. anonymous

See, take Log both the sides you will get: $\large Log(5^{x-3}.3^{2x-8}) = Log(225)$ Now use the formulas I have written above..

23. anonymous

Use the first formula only...

24. anonymous

$\log(5^{x-3}) . \log(3^{2x-8}) = \log 225$

25. anonymous

Now Check my first formula carefully...

26. anonymous

@pratu043 trust take my answer

27. anonymous

like what waterineyes did now

28. anonymous

That's what I've done here right @waterineyes?

29. anonymous

I give you an example: $\large Log(2 \times 3) = Log(2) + Log(3)$ Go by this procedure...

30. anonymous

I've already done that. So now we should use the second one?

31. anonymous

You have done it wrongly my friend...

32. anonymous

See check the formula.. After formula the result will be in sum form and your result is not in sum form.. Is there any + sign in your result???

33. anonymous

Oops. $\log(5^{x-3}) + \log(3^{2x-8}) = \log225$

34. anonymous

Yes now you are right... Now 225 is square of what number???

35. anonymous

15

36. anonymous

So can you write 225 as $$15^2$$ on right hand side.?? If yes then do it..

37. anonymous

$\log(5^{x-3}) + \log(3^{2x-8}) = \log15^{2}$

38. anonymous

Now use the power rule of Logarithms that is the Second Formula that I have given above.. Can you use that??

39. anonymous

I give you an example of that too: $\large Log(x)^4 = 4.Log(x)$

40. anonymous

$(x-3)\log5 + (2x-8)\log3 = 2\log15$

41. anonymous

Can you write 15 as 3*5??

42. anonymous

Yes. $(x-3)\log5 + (2x-8)\log3 = 2\log(5 * 3)$

43. anonymous

Then use it and use the first formula again on right hand side...

44. anonymous

$(x-3)\log5 + (2x-8)\log3 = 2\log5 + 2\log3$

45. anonymous

Now what you have to do is to compare the coefficients of $$Log3$$ and $$Log5$$ both the sides can you do that??

46. anonymous

Tell me the coefficient of Log3 on Left hand side and right hand side...

47. anonymous

On LHS its 2x - 8 and on RHS its 2.

48. anonymous

Just equate them and find x from it..

49. anonymous

Similarly equate the coefficients of log5 and then also find x you will get the same x in both the case...

50. anonymous

No no...

51. anonymous

$(2x-8)\log3 = 2\log3$

52. anonymous

I said to equate the coefficients buddy.. Coefficients are 2x - 8 and 2.. you should equate this only...

53. anonymous

So you should just do 2x-8 = 2?

54. anonymous

x = 5

55. anonymous

See, if : $\large xlog3 = ylog3$ implies x = y. Getting??

56. anonymous

Yes!!

57. anonymous

Yes you have found x = 5 that is absolutely right...

58. anonymous

Thanks a lot!!

59. anonymous

Now equate the coefficients of log5..

60. anonymous

x - 3 = 2 x = 2 + 3 = 5

61. anonymous

Yes got the same value of x so you are right.. If anyhow x values are coming different then either question is wrong or your solution is wrong.. Getting??