# Thread: Log equations - no solution?

1. ## Log equations - no solution?

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)

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
2 = log(x-9/x)
100 = x-9/x
100x = x-9
99x = -9
Same thing with this one.

2. Solve for x and v and then check the domain of log.

3. Your answers for v and x will give you a negative value, what is a log's domain?

4. Your workings are fine. However, it is correct that there are no solutions because of the domain restrictions.

$\displaystyle f(a) = \log[a]$ is only valid for [/tex]a>0[/tex]

In your first equation $\displaystyle v-2 > 0 \implies v > 2$

In your second equation $\displaystyle x-9 > 0 \implies x > -9$

If you check your solution against the domain restriction you'll find the values are not in the domain so there are no solutions

5. Thank you.

6. 1