WHAT'S GOING ON

>log(x^2) is not always equal to 2log(x)
This can't be true. There's no value for x to make this statement true.

>>7961486
SEARCH UP PROPERTIES OF LOGS AND DO THE SHIT YOURSELF

If you want all the neat exponentiation/logarithm rules to stay in effect, try to keep numbers involved non-negative. Once you consider all reals as bases/exponents/powers, many if not most neat rules will stop being true (and you will need complex numbers too).

>>7961486
>assuming x>0

Domain of ln(x) is x>0, but you can have negative x's provided that you square or take the absolute value of the argument. Therefore, ln(x^2) is not always the same as 2ln(x).

Pretty much ln(x^2) can take negative x-values, while 2ln(x) can't.

>>7961512
IT'S IMPOSSIBLE

>>7961518
NEGATIVE VALUES ALWAYS WORK WITH log(x^2) AND 2log(x)

>>7961486
2log(x) only has values of x > 0, while log(x^2) has values for all reals except 0.

Hence the "not always."

>>7961528
ln(0) is negative infinity. Negative infinity equals negative infinity.

>>7961508
One's undefined for x=-1, one's zero for x=-1.

They're equal on the domain of 2log(x), not for all reals.

>>7961532
Cool, then include zero as well.

>>7961536
Does that mean Symbolab is wrong?

File: untitled.png (21KB, 693x400px) Image search: [Google]

21KB, 693x400px

>>7961524
>Domain of ln(x) is x>0

Lel noob. Once you go complex, anything goes.

https://www.wolframalpha.com/input/?i=%28ln%28-2%29%29^2+-+2*ln%28-2%29

>>7961551
Not as much wrong, as apparently making assumptions (such as x>0 etc.) which other sources don't necessarily.

>>7961552

>>7961551

Why does log((-3)^2) work while 2log(-3) doesn't? Doesn't the log power rule say they're equivalent?

>>7961576
>>7961562

>>7961620
>>7961515

>>7961620
They are still equivalent when exponentiated thanks to Euler's formula.

It just gets wonky when you stay in the domain of logarithms, since the general rules are only true for certain for x>0

>>7961644
>the general rules are only true for certain for x>0
/thread

>>7961620
The arg of the logarithms has to be positive.
By using the logarithm laws you change the Df,

lg x^2 turned into 2lg x will also make the Df change since logarithms are only defined for x > 0.

>>7961674
in other words, i just write the absolute value of the argument whenever i jump using the log laws. but keep in mind that the domain of the function can change, and tell your examinatior how it is changed everytime you do that, otherwise all points lost on exam.

>>7961526
>NEGATIVE VALUES ALWAYS WORK WITH log(x^2) AND 2log(x)

It does not, negative values for 2log(x) is undefined.

[eqn]
\begin{align}
\log((x+i y)^2) &= \log(x^2+y^2) + i \arg((x+i y)^2) \\
2 \log(x+i y) &= \log(x^2+y^2) + i 2 \arg(x+i y) \\
\end{align}
[/eqn]
The problem is when the twice the argument of x+i y exceeds the range of arg.
For example
[eqn]
2 \arg(e^{i \displaystyle \tfrac{3 \pi}{4}}) = 2 \frac{3 \pi}{4} = \frac{3 \pi}{2} \\
\arg( ( e^{i \displaystyle \tfrac{3 \pi}{4}} )^2) = \arg( e^{i \displaystyle \tfrac{3 \pi}{2}}) = \arg(e^{i \pi} e^{i \displaystyle \tfrac{\pi}{2}}) = -\frac{\pi}{2}
[/eqn]

File: untitled.png (41KB, 673x849px) Image search: [Google]

41KB, 673x849px

>>7961689
>negative values for 2log(x) is undefined

How so?

File: marvin-gaye-02.jpg (887KB, 2616x1470px) Image search: [Google]

887KB, 2616x1470px

>>7961486
OOO YEAH
WHAT'S GOING ON

>>7961738
Im sorry, i did not mean undefined.

>>7961524
r · e^θi

File: not_defined_for_negative_real_numbers.png (9KB, 665x229px) Image search: [Google]

9KB, 665x229px

>>7961738
>>7961745
Taking it back,
did not see that you had a complex value plot in your graph.

Like I said, for all real number x
all logarithms are only defined for x > 0.

Therefore, 2log(x) is not defined for negative numbers while the other one is.

pic related, real-valued plot.