Why Do These Log Equations Have No Solutions?

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SUMMARY

The discussion centers on the lack of solutions for the logarithmic equations log(v-2) = 1 + log(v+2) and 2 + log(x) = log(x-9). The first equation simplifies to v = -22/9, which results in a negative argument for the logarithm, thus confirming no solution exists. Similarly, the second equation leads to x = -11, where both log(x) and log(x-9) are undefined due to negative inputs. Therefore, both equations have no valid solutions.

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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.
 
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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!)
 
TN17 said:

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.
 

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