## anonymous 5 years ago condense the expression: 1/2Log5 16-3log5 x+4Log5y

1. anonymous

first move the coefficients that are infront of the logarithms to the right

2. anonymous

meaning log$\log_{5} 16^{1/2}$

3. anonymous

ignore the first log my equations button is not working

4. anonymous

but the log expression below is correct use that example to simplify the other two

5. anonymous

then let me know what you got

6. anonymous

log5y^x+4?

7. anonymous

$\log_{5} 4-\log_{5} x ^{3}+\log_{5} y ^{4}$

8. anonymous

thankyou for helping me their just really hard for me ;(

9. anonymous

no problem

10. anonymous

have time for more?

11. anonymous

we aren't done yet sorry, now when you are subtract two logs with the same base do you condense them by multiplying them together or dividing them?

12. anonymous

dividing

13. anonymous

yes so log base 4 is on top the numerator making the other two...?

14. anonymous

on the bottom being multiplied together

15. anonymous

idk

16. anonymous

17. anonymous

18. anonymous

4/x^3+y^4

19. anonymous

when you are adding two logs together with the same base then you multiply them together

20. anonymous

okay

21. anonymous

$\log_{5} (4/x ^{3}y ^{4})$

22. anonymous

does that make sense

23. anonymous

yes

24. anonymous

i just want to make sure you understand rather than just giving you the answer so that later you can ace a test on this if you need to

25. anonymous

yah this good

26. anonymous

ok why we try another problem. this will be the last one i can help with, then i have to go alright?

27. anonymous

okay

28. anonymous

solve the equation; 15+2 log2 x=31

29. anonymous

so when solving for any variable the whole point is to get the variable by itself so what would be one thing we can move away from the side with x easily?

30. anonymous

yes

31. anonymous

that was a question

32. anonymous

33. anonymous

woudnt it be 17?

34. anonymous

i didn't mean the final answer, i meant that it will be easy to move the fifteen since the fifteen is not attached to anything

35. anonymous

move to be 15

36. anonymous

yes we subtract fifteen from both sides then we simplify the logarithm but first what does the new equation look like?

37. anonymous

it will equeal 16instead of 31

38. anonymous

yes

39. anonymous

know we focus on the log

40. anonymous

;)

41. anonymous

okay!

42. anonymous

what can we move to condense the logarithm more?

43. anonymous

the 2 in front

44. anonymous

yes

45. anonymous

ah yes!!

46. anonymous

so what do we have now?

47. anonymous

logx=16

48. anonymous

no... $\log_{2} x ^{2}=16$

49. anonymous

does that make sense?

50. anonymous

yes

51. anonymous

okay a trick that i learned is you put the same base of the log you want to remove under both sides of the equation like so: $2^{\log_{2}x ^{2} }=2^{16}$

52. anonymous

now what can you simplify?

53. anonymous

exponents

54. anonymous

yes $x ^{2}=2^{16}$ is our new equation as 2^log base 2 equals one

55. anonymous

now what would be the next step?

56. anonymous

logx=4

57. anonymous

why would we do that when we already have x by itself. what we would do is simplify 2^16=65536

58. anonymous

then x^2=65536

59. anonymous

65536?

60. anonymous

yes now what would you do to get x all by itself?

61. anonymous

take the root of x and the other side

62. anonymous

x=256

63. anonymous

does that make sense?

64. anonymous

thats what i thought

65. anonymous

$\sqrt{x ^{2}}=\sqrt{65536}=x=256$

66. anonymous

omg thankyou so much

67. anonymous

no problem I have to go now, but i wish you the best of luck :)

68. anonymous

thankyou