Why Do These Log Equations Have No Solutions?

[TN17]

Homework Statement

In the answers, it says these two don't have any solutions. I don't understand how, though.

log(v-2) = 1 +log(v+2)
and
2 + logx = log(x-9)

The Attempt at a Solution

For the first one, I did:
log(v-2/v+2) = 1
v-2/v+2 = 10
v-2 = 10v + 20
0 = 9v + 22
So I didn't get no solution. I don't know if this is right, though.

For the second one, I did:
2 = log(x-9) - logx
10² = log(x-9/x)
100 = x-9/x
100x = x-9
99x = -9
Same thing with this one.

 

[rock.freak667]

You are going correct, now when you find 'v' and 'x'. Put them back into the original equations and see if you can actually calculate it.

(remember, the log of a negative number does not exist!)

 

[HallsofIvy]

Homework Statement

In the answers, it says these two don't have any solutions. I don't understand how, though.

log(v-2) = 1 +log(v+2)
and
2 + logx = log(x-9)

The Attempt at a Solution

For the first one, I did:
log(v-2/v+2) = 1
v-2/v+2 = 10
v-2 = 10v + 20
0 = 9v + 22
So I didn't get no solution. I don't know if this is right, though.

"I didn't get no solution" grammatically correct! I love it!
You also did not finish the problem: v= -22/9 so that v- 2= -22/9- 2< 0. That is NOT a solution because log is not defined for negative numbers.

For the second one, I did:
2 = log(x-9) - logx
10² = log(x-9/x)
100 = x-9/x
100x = x-9
99x = -9
Same thing with this one.

So x= -11. Neither log x nor log (x- 9) is defined.