Two ways to solve log problem.

Hi,

I show two workings to solve the equation below using the simplification rules for logs.

My textbook does not indicate how and when solutions are lost using the log simplification rules. Clearly if we use the $\displaystyle \log_b (xy) = \log_b x + \log_b y$ rule negative solutions might be lost.

Where can I find more information about this I want to understand the problem better.

Given $\displaystyle \log_2 (x^2) = 4$:

Either:

$\displaystyle \log_2 (x^2) = 4$

$\displaystyle \log_2 (x) + \log_2 (x) = 4$

$\displaystyle 2\log_2 (x) = 4$

$\displaystyle \log_2 (x) = 2$, divide both sides by $\displaystyle 2$

$\displaystyle x = 2^2$

$\displaystyle x = 4$

Or:

$\displaystyle \log_2 (x^2) = 4$

$\displaystyle 2^4 = x^2$

$\displaystyle x^2 = 16$

$\displaystyle x = \pm 4$

The first solution is wrong. The rule doesn't hold (looses solutions) for even exponents.