# solution to equation with logs

• Sep 20th 2009, 12:06 AM
eniuqvw
solution to equation with logs
I want to know why log(9x^2) = 0 has two solutions. If we write it as log(9) + log(x^2) = 0, ==> log(9) + 2 log(x) = 0 ==> 2log(x) = -log(9) ==> log(x) =-[log(9)]/2 ==> x = 10 ^ ([-log(9)]/2), there is nothing in the algebraic manipulation that would suggest two solutions because each move was via some rule of common logs. Of course, i understand intuitively why it has two solutions, but why does the algebraic manipulation allow us to arrive at only one solution?
• Sep 20th 2009, 12:10 AM
mr fantastic
Quote:

Originally Posted by eniuqvw
I want to know why log(9x^2) = 0 has two solutions. If we write it as log(9) + log(x^2) = 0, ==> log(9) + 2 log(x) = 0 ==> 2log(x) = -log(9) ==> log(x) =-[log(9)]/2 ==> x = 10 ^ ([-log(9)]/2), there is nothing in the algebraic manipulation that would suggest two solutions because each move was via some rule of common logs. Of course, i understand intuitively why it has two solutions, but why does the algebraic manipulation allow us to arrive at only one solution?

$\log x^2 \neq 2 \log x$.

$\log x^2 = 2 \log {\color{red}|} x{\color{red}|}$. Two solutions follow from this.

The easier approach to the question is to solve $9 x^2 = 1$.
• Sep 20th 2009, 12:17 AM
eniuqvw
oops
ok, i should have realized that the rule log(b^t) = t [log(b)] is defined for positive b, whereas in log(9x^2), x can take any real value. Thanks for the help