logarithm sum plz help?

Question

ok heres the sum

2log base3 y - log base3 (y+4) = log base3 9

well all i knw is that ''log base 3 9'' = 2

i been trying this sum 4 abt 2 hours and i couldnt get it can someone solve it 4 me and give me some tips or anything usefeull

thanks

Answer 1

2log base3 y - log base3 (y+4) = log base3 9

log base3 y² - log base3 (y+4) = log base3 9

log base3 (y²/(y+4)) = log base3 9

Since the base of both the logarithms is same i.e. 3

y²/(y+4) = 9

y² = 9(y+4)

y² -9y- 36 =0

y² -12y+ 3y- 36 =0

(y-12)(y+3) = 0

y = 12,-3

y cannot be negative ,discard y= -3 (since log of a negative number is nonexistent)

Thus y =12

Answer 2

2log base3 y = log base3 y^2
log base3 y^2 - log base3 (y+4) = log base3 [(y^2)/ (y+4)] = log base3 9
(y^2)/(y+4) = 9
y^2 - 9y - 36 = 0
(y - 12)(y + 3)
y = 12 or y = -3
but y should be greater than zero,
therefore, y = 12

Answer 3

2log base3 y-log base3(y+4)= log base3 (y2/(y+4))=log base3 9
log base3 9=2=log base3(y2/(y+4))=>3^2=y2/(y+4)
9=y2/(y+4)=> 9y+36=y2=>y2-9y-36=0 than factor
(y-12)(y+3)=0 => y=12, -3

Answer 4

2log y - log (y+4) = log 9 ---------- all base3

log y^2 - log (y+4) = log 9

log {y^2/ (y+4)} = log 9

{y^2/ (y+4)}= 9

y^2 = 9 (y+4)

y^2 - 9 (y+4) = 0

y^2 - 9 y - 36 = 0

y^2 - 12y+ 3y - 36 = 0

(y - 12) (y+3) =0

y = 12 0r - 3.
----------------------------------------------------------------------------

If you know log (base3) 9 = 2

Proceed like this.

2log y - log (y+4) = 2 . ---------- all base3
log y - (1/2) log (y+4) = 1
log y - log (y+4)^ (1/2)= 1
log {y/ (y+4)^ (1/2)} = log 3 since log (base 3) 3 = 1
{y/ (y+4)^ (1/2)} = 3

y = 3 (y+4)^ (1/2)

Squaring y^2 = 9 (y+4)
We get the same equation as before.

Answer 5

all logs to base 3
2logy-log[y+4]=log9
log[y^2/y+4]=log9
y^2=9[Y+4]
Y^2-9Y-36=0
Y=[9+-sqrt[81+4*36]]/2
=[9+3sqrt[9+16]]/2
=[9+-15]/2
=12,-3