How to solve for x? (logarithm)

[nigahiga]

3 log_10(x^2)-log_10^2(-x) = 9
Not really sure how to deal with the log(-x). What can I do with this equation?

 

[Deleted member 4993]

3 log_10(x^2)-log_10^2(-x) = 9
Not really sure how to deal with the log(-x). What can I do with this equation?

substitute

u = -x

Now you have:


3 log_10(u^2) - log_10^2(u) = 9

and continue....

 

[nigahiga]

Also, when dealing with logarithms, is checking for extraneous solutions essential? What constitutes that a certain logarithmic equation will not yield extraneous solution? Or is it pure arbitrary?

 

[Deleted member 4993]

Also, when dealing with logarithms, is checking for extraneous solutions essential? What constitutes that a certain logarithmic equation will not yield extraneous solution? Or is it pure arbitrary?

Log(-a) when a ≥ 0 does not exist in real domain.

However, if a < 0 then log(-a) and log(a²) exist in real domain.

 

[nigahiga]

substitute

u = -x

Now you have:


3 log_10(u^2) - log_10^2(u) = 9

and continue....

What happens to the first term when you substitute u = -x? Does the 3 log_10(x^2) ---> 3 log_10((-u)^2) ---> 3 log_10(u^2) ?

 

[Deleted member 4993]

What happens to the first term when you substitute u = -x? Does the 3 log_10(x^2) ---> 3 log_10((-u)^2) ---> 3 log_10(u^2) ?

Yes

 

[nigahiga]

Yay it worked! Thank you very much

Question regarding this method. If -x can be substituted as u, then 3 log_10(x^2) can be written as:

3 log_10(-u^2) ---> 3 log_10(u^2) ---> 6 log_10(u) ?
but on the first step, if I move the exponent down earlier, then

3 log_10(-u^2) ---> 6 log_10(-u)

and 6 log_10(-u) ≠ 6 log_10(u)

So why does this method of squaring the -u work out in the end? Is the domain at work here?