Doubt on logarithms

[DyuHaf23]

There's a very well known rule that says that
[math]log_a (b^n) = nlog_ab[/math]
But I've come across something that has me a bit confused. Consider this equation:
[math]ln(x)^2 = 1[/math]If you solve it using the classing method of exponentiating both sides of the equation, you get
[math]x^2 = e[/math]which has solutions for x equals to [imath]-\sqrt(e)[/imath] and [imath]\sqrt(e)[/imath]
But if you use the aforementioned rule and take the exponent out of the logarithm, you get
[math]2lnx = 1[/math][math]lnx = 1/2[/math][math]x = \sqrt(e)[/math]Which doesn't include part of the solution. What am I missing here?

 

[Steven G]

That well known theorem that you stated is not true, ok, it's incomplete.

For example \(\displaystyle \log{(-3)}^2 \neq 2\log{3}\). Why not??


Can you state the complete well known theorem.
Your teacher and/or textbook is not doing a good job if you think that the theorem you stated is correct?


For completeness note that log[(-2)(-3)] ≠ log(-2)log(-3) !!!!

 

[Deleted member 4993]

There's a very well known rule that says that
[math]log_a (b^n) = nlog_ab[/math]
But I've come across something that has me a bit confused. Consider this equation:
[math]ln(x)^2 = 1[/math]If you solve it using the classing method of exponentiating both sides of the equation, you get
[math]x^2 = e[/math]which has solutions for x equals to [imath]-\sqrt(e)[/imath] and [imath]\sqrt(e)[/imath]
But if you use the aforementioned rule and take the exponent out of the logarithm, you get
[math]2lnx = 1[/math][math]lnx = 1/2[/math][math]x = \sqrt(e)[/math]Which doesn't include part of the solution. What am I missing here?

Logarithm of negative numbers cannot be defined in real domain. Thus for real domain, x = -\(\displaystyle \sqrt{e}\) is an extraneous solution.

 

[pka]

There's a very well known rule that says that
[math]log_a (b^n) = nlog_ab[/math] Correct

But I've come across something that has me a bit confused. Consider this equation:
[math]ln(x)^2 = 1[/math]If you solve it using the classing method of exponentiating both sides of the equation, you get

BUT [imath]\log(x)^2 [/imath] is confused notation.
Does that mean [imath]\log(x^2)\text{ or }\log^2(x) [/imath] Those two are different!
You also learn the change of base theorem: [imath]\log_a(b)=\dfrac{\log(b)}{\log(a)}[/imath]

 

[greg1313]

[math]log_a (b^n) = nlog_ab[/math]
[math]ln(x)^2 = 1[/math]

Look at the placement of the grouping symbols in that rule.

Don't change the order when you write your example:
\(\displaystyle ln(x^2) \ = \ 1.\)

\(\displaystyle b^n \ \ and \ \ x^2 \ \) are meant to be the arguments, respectively.

 

[DyuHaf23]

That well known theorem that you stated is not true, ok, it's incomplete.

For example \(\displaystyle \log{(-3)}^2 \neq 2\log{3}\). Why not??


Can you state the complete well known theorem.
Your teacher and/or textbook is not doing a good job if you think that the theorem you stated is correct?


For completeness note that log[(-2)(-3)] ≠ log(-2)log(-3) !!!!

The theorem also states that the argument of the function has to be greater than 0. Is that the issue? Does that mean this property can't be used if you're not sure the argument is greater than 0?
Edit: obviously the argument has to be greater than 0 in the logarithm. What I meant was the base (the b in [imath]b^n[/imath]). I think the problem is you can't just assume that the b is positive, so you can't apply this theorem here. This doubt actually came up when I was trying to plot the graph of a function with a similar logarithm, and I assumed that it wasn't defined for values lower than 0, and then it turned out I was wrong.