log(9y^2+19y+2)-log(y+2)=2

Question

I am so bad at logs need help please

Answer 1

We have here a difference of logs.  According to the properties of logarithms, we can write this as a quotient.

 

log((9y² - 19y + 2) / (y + 2)) = 2

 

 

The solution to a logarithm is the exponent of base number 10.

 

(9y² - 19y + 2) / (y + 2) = 10²

 

(9y² - 19y + 2) / (y + 2) = 100

 

 

Next, we factor out the numerator using FOIL, so we can hopefully cancel out the (y + 2) term. 

 

((9y + 1) (y + 2)) / (y + 2) = 100

 

 

We are able to factor out.

 

9y + 1 = 100

 

Now, the equation can be easily solved.

 

9y = 99

  y = 11

 

Answer 2

log(9y²+19y+2)-log(y+2) = log[(9y²+19y+2)/(y+2)]

 

Since 10^log(x)=x, the original equation could be

 

10^{log[(9y2+19y+2)/(y+2)]} = 10²

 

which is

 

9y²+19y+2 = 200y+200

 

Then

 

9y²-81y-198 = 0

 

Divide by 9

 

y²-9y-22 = 0

 

which can be written as

 

(y-11)(y+2) = 0

 

y = 11 or y = -2

 

since log(x) is not defined at x = 0, so y = -2 is abandoned

 

Therefore, the correct answer is 

 

y = 11