pratu043 3 years ago Can someone please explain how to solve a log equation with an example.

1. dpaInc

how 'bout u pick an example....

2. dpaInc

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

3. pratu043

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

4. pratu043

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

5. pratu043

I mean new to log.

6. pratu043

pls help.

7. lgbasallote

welcome to logarithms!

8. pratu043

Yeah thanks!

9. waterineyes

I guess you can solve this without using Logs..

10. pratu043

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

11. waterineyes

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

12. moha_10

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

13. pratu043

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

14. moha_10

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

15. moha_10

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

16. pratu043

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

17. waterineyes

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

18. moha_10

before we take the logarithem

19. moha_10

it's from exponent experepties

20. pratu043

Why did you delete that @dpaInc?

21. dpaInc

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

22. waterineyes

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. waterineyes

Use the first formula only...

24. pratu043

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

25. waterineyes

Now Check my first formula carefully...

26. moha_10

27. moha_10

like what waterineyes did now

28. pratu043

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

29. waterineyes

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

30. pratu043

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

31. waterineyes

You have done it wrongly my friend...

32. waterineyes

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. pratu043

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

34. waterineyes

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

35. pratu043

15

36. waterineyes

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

37. pratu043

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

38. waterineyes

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

39. waterineyes

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

40. pratu043

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

41. waterineyes

Can you write 15 as 3*5??

42. pratu043

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

43. waterineyes

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

44. pratu043

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

45. waterineyes

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

46. waterineyes

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

47. pratu043

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

48. waterineyes

Just equate them and find x from it..

49. waterineyes

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

50. waterineyes

No no...

51. pratu043

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

52. waterineyes

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

53. pratu043

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

54. pratu043

x = 5

55. waterineyes

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

56. pratu043

Yes!!

57. waterineyes

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

58. pratu043

Thanks a lot!!

59. waterineyes

Now equate the coefficients of log5..

60. pratu043

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

61. waterineyes

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??