## taranicolee Group Title solve log5=log4+log(-4x+5) 2 years ago 2 years ago

1. JamesJ Group Title

first simplify the RHS (right hand side), what do you get?

2. Mr.crazzy Group Title

x=1.225

3. taranicolee Group Title

well it will equal 5=4+(-4x+5) and you would solve from there, but OH WAIT. haha for some reason when i was trying to solve it i wrote 4 times (-4x+5)

4. JamesJ Group Title

@crazzy: it's actually almost never helpful just to give an answer without explaining how you got the answer. But now that you've given the answer, you can now explain step by step how you got there. :-)

5. taranicolee Group Title

could you help me with another problem? $9^{x-5} = 81^{3}$

6. taranicolee Group Title

waiiit.. james j :/ hahah

7. myininaya Group Title

$\log(ab)=\log(a)+\log(b)$

8. myininaya Group Title

Do you recall this?

9. taranicolee Group Title

yes

10. myininaya Group Title

$\log(5)=\log(4(-4x+5))$

11. myininaya Group Title

so now we can set the insides equal

12. myininaya Group Title

$5=4(-4x+5)$

13. taranicolee Group Title

THAT'S WHY I HAD IT MULTIPLYING. haha sweet (:

14. sasogeek Group Title

15. myininaya Group Title

$5=-16x+20$ I distributed the 4 here

16. myininaya Group Title

Now I'm going to subtract 20 on both sides $5-20=-16x$

17. myininaya Group Title

$-15=-16x => -16x=-15$

18. myininaya Group Title

Now we divide both sides by -16 $x=\frac{-15}{-16}$

19. myininaya Group Title

But a negative divided by a negative is positive

20. myininaya Group Title

$x=\frac{15}{16}$

21. sasogeek Group Title

what did i do wrong? my answer doesn't concur with yours myininaya... :|

22. taranicolee Group Title

i think that's right though because when i tried to solve it on my own i got one, and 15/16 .9375

23. sasogeek Group Title

i know my working is wrong, i don't see where i went wrong though...

24. taranicolee Group Title

myininaya could you help me with the other equation i put up there please?

25. myininaya Group Title

@saso I can't read yours that well

26. taranicolee Group Title

sasogeek, you didn't just cancle out the log, you cancled the five as well, right?

27. Mr.crazzy Group Title

28. sasogeek Group Title

log5-log5=0

29. myininaya Group Title

Taranicolee you can post if you wish

30. taranicolee Group Title

sasogeek you have to distribute the four before you subtract from in the parenthesis

31. taranicolee Group Title

9 x−5 =81 3

32. taranicolee Group Title

oh wait that's not it

33. myininaya Group Title

by the way $\log(4x+5) \neq \log(4x)+\log(5)$

34. taranicolee Group Title

$9^{x -5}=81^{3}$

35. sasogeek Group Title

i'm just noticing that, thanks for pointing it out. i forgot the log(ab)=log(a)+log(b) rule :)

36. myininaya Group Title

$81=9^2$ agree?

37. taranicolee Group Title

yes

38. myininaya Group Title

$9^{x-5}=(9^2)^3$

39. myininaya Group Title

so i just replaced 81 with 9 squared

40. taranicolee Group Title

ohh to get the bases the same so you can cancle out right?

41. myininaya Group Title

now law of exponents tells us $(x^a)^b=x^{a \cdot b}$

42. myininaya Group Title

if you get the same base on both sides then the exponents have to be equal in order for the equation to hold

43. myininaya Group Title

$9^{x-5}=9^6 =>x-5=6$

44. taranicolee Group Title

you multiplied the exponents on the right side of the equation right ? then cancled out the bases and now to solve you just add five to 6?

45. myininaya Group Title

do you recall the law of exponent i mentioned?

46. taranicolee Group Title

no :/

47. myininaya Group Title

$(x^a)^b=x^{a \cdot b}$?

48. taranicolee Group Title

nope. i wasn't here for the teaching of this

49. myininaya Group Title

yes if is it in the form you just multiply the exponents to simplify

50. taranicolee Group Title

*there

51. myininaya Group Title

so for example if we have $(3^2)^3=3^{2 \cdot 3} =3^6$

52. myininaya Group Title

or if we have $(43^4)^3=43^{4 \cdot 3} =43^{12}$

53. taranicolee Group Title

oh sweet ! you're awesome. wanna move to italy and be my tutor?? hahah. if the equation were $16^{x}=8$ not in that form, how would you solve?

54. myininaya Group Title

I would love to go to Italy. write both sides with same base both 16 and 8 can be written as 2 to the power of something

55. taranicolee Group Title

2^4 and 2^3

56. myininaya Group Title

yes! :)

57. myininaya Group Title

$(2^4)^x=2^3$ so this is our equation

58. taranicolee Group Title

cancle bases, 4x=3 divide by 4 on both sides?

59. myininaya Group Title

perfect! :)

60. taranicolee Group Title

YAYYYY. thank you !!!!

61. myininaya Group Title

You're welcome.

62. myininaya Group Title

Thank you for being eager to learn this stuff.

63. taranicolee Group Title

((: