Two ways to solve log problem.

• May 2nd 2009, 12:05 AM
craigmain
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 $\log_b (xy) = \log_b x + \log_b y$ rule negative solutions might be lost.

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

Either:

$\log_2 (x^2) = 4$
$\log_2 (x) + \log_2 (x) = 4$
$2\log_2 (x) = 4$
$\log_2 (x) = 2$, divide both sides by $2$
$x = 2^2$
$x = 4$

Or:

$\log_2 (x^2) = 4$
$2^4 = x^2$
$x^2 = 16$
$x = \pm 4$

The first solution is wrong. The rule doesn't hold (looses solutions) for even exponents.
• May 2nd 2009, 01:06 AM
chisigma
The reason of the [apparent] contradiction is that the identity $\log_{2} x^{2}= 2\cdot \log_{2} x$ is true only if $x>0$...

Kind regards

$\chi$ $\sigma$
• May 2nd 2009, 01:21 AM
craigmain
Hi,

What does that mean for the multiplicative rule.

$\log_b x y = \log_b x + \log_b y$

Clearly for this rule, $x \neq y$ if $x < 0$
• May 2nd 2009, 02:02 AM
chisigma
My previous answers has been a little 'approximated' and I apologize for that(Worried) ...

If x is a real number and x<0, then is...

$\log_{2} x = \frac{\ln x}{\ln 2}= \frac{\ln |x| + i\cdot \pi}{\ln 2}$

In this case however is

$\log_{2} x^{2} = 2\cdot \frac{\ln x}{\ln 2}= \frac{2\cdot \ln |x| + i\cdot2\cdot \pi}{\ln 2}= \frac{2\cdot \ln |x|}{\ln 2}$

The consequence is that the general expression you have written holds for any value of b, x, and y with $b>0$, $x \ne 0$ and $y \ne 0$...

Kind regards

$\chi$ $\sigma$