Two ways to solve log problem.

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May 1st 2009, 11:05 PM
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.

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

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, 12:06 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, 12:21 AM
craigmain

additionally,

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, 01:02 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

$\chi$ $\sigma$