Indices and Logarithms (1 Viewer)

mathsbrain · Jun 1, 2017

hi, is it safe to assume index laws and logarithmic laws only apply(apply without need to confirm other things...) for positive bases?
E.g. for equations like log(base x) 4=2, we only get the answer of 2 and not negative 2 right?

InteGrand · Jun 1, 2017

mathsbrain said:
hi, is it safe to assume index laws and logarithmic laws only apply(apply without need to confirm other things...) for positive bases?
E.g. for equations like log(base x) 4=2, we only get the answer of 2 and not negative 2 right?

Log based are generally restricted to be positive numbers (excluding 1), so yes.

mathsbrain · Jun 1, 2017

InteGrand said:
Log based are generally restricted to be positive numbers (excluding 1), so yes.

thanks, what's the reason behind this?

InteGrand · Jun 1, 2017

mathsbrain said:
thanks, what's the reason behind this?

Basically log_b(x) is the inverse function to b^x when b > 0 (and b ≠ 1), but if b < 0, then b^x isn't generally well-defined in the real numbers for general real x (like (-1)^1/2 doesn't exist in the real numbers (and in fact is multivalued in complex numbers)), so we can't define an inverse function to it.

$\noindent Another way to say it is say we wanted to define log base $-2$, and we wrote $\log _{-2} (y) = x$. This should mean $y = (-2)^{x}$, but since powers of negative numbers are generally undefined in the reals (unless the $x$ here is an integer), the only values for $y$ for which we could get a sensible result would be ones where the $x$ is a rational number with odd denominator, so we can't get a $\log_{-2}(y)$ for all positive $y$. So it would not as useful as the logs with positive bases $b$ (with $b \neq 1$).$

mathsbrain · Jun 2, 2017

InteGrand said:
Basically log_b(x) is the inverse function to b^x when b > 0 (and b ≠ 1), but if b < 0, then b^x isn't generally well-defined in the real numbers for general real x (like (-1)^1/2 doesn't exist in the real numbers (and in fact is multivalued in complex numbers)), so we can't define an inverse function to it.

$\noindent Another way to say it is say we wanted to define log base $-2$, and we wrote $\log _{-2} (y) = x$. This should mean $y = (-2)^{x}$, but since powers of negative numbers are generally undefined in the reals (unless the $x$ here is an integer), the only values for $y$ for which we could get a sensible result would be ones where the $x$ is an integer, so we can't get a $\log_{-2}(y)$ for anything other than $y$ of the form $(-2)^{n}$ (where $n$ is an integer). So it would not as useful as the logs with positive bases $b$ (with $b \neq 1$).$

wow very advanced haha, thanks! more reading now!
but generally speaking for INDICES a^x, the base a CAN be negative right? So for example, x^(2/3)=4, we can have -8?

I guess what i'm trying to clarify is, in logs, base cant be negative, but in exponentials it can? And thats why in the example x^2/3=4, we have to answer +/-8 and not just 8??

1729 · Jun 2, 2017

mathsbrain said:
wow very advanced haha, thanks! more reading now!
but generally speaking for INDICES a^x, the base a CAN be negative right? So for example, x^(2/3)=4, we can have -8?

I guess what i'm trying to clarify is, in logs, base cant be negative, but in exponentials it can? And thats why in the example x^2/3=4, we have to answer +/-8 and not just 8??

$$\noindent Yes, $-8$ is a solution because the equation can be written as $x^2 = 64$. If $b \leq 0$ or $b=1$ there would be few values of $x$ that satisfies $b^y = x$ (ie. $\log_bx = y)$ and so logarithms with base $b\leq0$ or $b=1$ are not as useful as those with bases $b>0$ $(b\neq1).$

Indices and Logarithms (1 Viewer)

mathsbrain

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InteGrand

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mathsbrain

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