Did I solve this log problem right?

Question

If I didn't please clarify those steps I did wrong

Problem:

log y+ log (y+21) =2

Solving:

log(base 10) y (y+21) = log (base 10) 10 ^2
y (y+21) =100
y^2 +21y -100 = 0

How do I factor it out? I used the quadratic formula but it still doesn't work

Answer 1

log(y) + log(y + 21) = 2

Combine the logs,

log [ y(y + 21) ] = 2

log [ y^2 + 21y ] = 2

Exponential form,

10^2 = y^2 + 21y

100 = y^2 + 21y

0 = y^2 + 21y - 100

This factors as

(y + 25)(y - 4) = 0

This means y = { -25, 4 }

However, y = -25 is an extraneous solution (since plugging it into the original results in taking the log of a negative number, which is strictly not allowed). Therefore, we reject -25 and

y = 4

Answer 2

You're doing fine.
When you factor, you want to come up with two numbers that you can multiply together to get the last number, and add together to get the middle number. A lot of people have problems with factoring, at first, it takes practice.
If you want to use the quad formula, that works too. You want y = [-21 +/- sqrt (21^2 - 4x1x(-100))] / 2(1) =
[-21 +/- sqrt (441 + 400)]/2 =
[-21 +/- 29]/2 =
-50/2 or 8/2 that is
-25 or 4. Since log arguements have to be positive, throw away the -25 and keep the 4
Check:
log (4 x (4+21)) = log (4x25) = log 100 = 2
As you see, the quad formula is more steps, therefore more chance of making mistakes. However, it's good for if the polynomial doesn't factor, or even if it does factor and you don't see the factors.

Answer 3

y(y+21)=100
y^2 +21y = 100
(y+21/2)^2 = 100 + (21/2)^2 = 841/4 = (29/2)^2
y+21/2 = +/- 29/2
y = 4 or -25
but question had log(y) and log(y+21), so select y=4

Answer 4

you're completely correct!

the factoring would be (y-4) (y +25)