Natural log is tricky

[YehiaMedhat]

My work: was combinig the 2 ln on the l.h.s so that
ln(a+x)=ln(-a/x-2), hence a+x = -a/x-2, then by canceling a for both sides and factorization x comes to equal 1
But this is not the answer

 

[Dr.Peterson]

My work: was combining the 2 ln on the l.h.s so that
ln(a+x)=ln(-a/x-2), hence a+x = -a/x-2, then by canceling a for both sides and factorization x comes to equal 1
But this is not the answer
View attachment 34501

It is not true that ln(a) + kn(x) = ln(a + x).

What should it be? Check the rules for logs!

But also, "canceling" a in a+x = -a/(x-2) does not leave you with x = -1/(x-2), which I assume is what you did. "Canceling" doesn't work that way.

Finally, if your answer is 1, you would pick the answer that is an interval containing 1; there is such an answer, so I'm not sure why you said 1 is not the answer.

 

[Steven G]

I will be generous and start you off:

ln(a)+ln(x) = ln[(-a)/(a-2)]
ln(ax)= ln[(-a)/(a-2)]
ax = -a/(a-2) under certain conditions. What are those conditions???
Now solve for x

 

[Dr.Peterson]

I will be generous and start you off:

ln(a)+ln(x) = ln[(-a)/(a-2)]
ln(ax)= ln[(-a)/(a-2)]
ax = -a/(a-2) under certain conditions. What are those conditions???
Now solve for x

You copied it wrong. Your version of the problem would be a little harder.

 

[pka]

This is a flawed question. Is it possible to have both [imath]\log(a)~\&~\log(-a)[/imath] in the same question?

[imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath]

 

[Steven G]

This is a flawed question. Is it possible to have both [imath]\log(a)~\&~\log(-a)[/imath] in the same question?

No, I disagree with that. yes, a>0 and -a/(x-2) = a/(2-x)>0. That is a>0 and x<(I prefer not to say). That should be fine, no?

Besides, the is no log(-a)!

 

[pka]

Besides, the is no log(-a)!

Are you serious? You don't think that
[imath]\log\dfrac{-a}{x-2}[/imath] is meant to be [imath]\log\left(\dfrac{-a}{x-2}\right)=\log(-a)-\log(x-2)[/imath]
[imath][/imath][imath][/imath]

 

[Steven G]

Are you serious? You don't think that
[imath]\log\dfrac{-a}{x-2}[/imath] is meant to be [imath]\log\left(\dfrac{-a}{x-2}\right)=\log(-a)-\log(x-2)[/imath]
[imath][/imath][imath][/imath]

No, I do not think blindly that log(-a/(x-2)) = log(-a) - log(x-2)

log(4) = log(-4/-1) = log(-4)-log(-1) = undefined. So log 4 is undefined, I don't think so.
Assuming a/b>0, log(a/b) = log a - log b iff a and b are positive

 

[topsquark]

Are you serious? You don't think that
[imath]\log\dfrac{-a}{x-2}[/imath] is meant to be [imath]\log\left(\dfrac{-a}{x-2}\right)=\log(-a)-\log(x-2)[/imath]
[imath][/imath][imath][/imath]

Order of operations: Do what's inside the parenthesis first!

-Dan