I am reviewing algebra 2 using some video tutorial from mathtutordvd.com.
In one of the videos, the author converts a logarithm equation into the exponent form as follows:
Which logarithm identity is used to convert the equation?
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That's just because the exponential function is well defined. – mwt Sep 26 '18 at 18:58
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They just took the exponential of both sides... if $x=y$ then $6^x=6^y$ – user438666 Sep 26 '18 at 19:03
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1This being said, it is far too complicated. Logarithms in any base are bijections from $\mathbf R_+$ to $\mathbf R$, so $\log_n(2x-3)=\log_b 4$ implies $2x-3=4$. – Bernard Sep 26 '18 at 19:03
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@Bernard That is only true if $n=b$. $\log_2 4= \log_3 9$ and $4\neq 9$. – Radial Arm Saw Jul 20 '20 at 16:07
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1@RadialArmSaw: I don't remember why I wrote such a thing, but you're right, $n$ should be $b$ in my comment. Probably one of my many lapsus calami. – Bernard Jul 20 '20 at 17:19
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@Bernard It happens to all of us. Lol – Radial Arm Saw Jul 20 '20 at 18:09
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We are using the definition
$$\log_a b=c\iff a^c=a^{\log_a b}=b$$
but since $\log$ function is injective we can conclude directly
$$\log_a f(x)=\log_a g(x)\iff f(x)=g(x)$$
for $f(x),g(x)>0$.
user
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This comes directly from the definition of a function, that for a certain input, a function produces a unique output. In other words, for a function $f$,
$$a=b \Rightarrow f(a)=f(b)$$
By definition, exponents are functions, that is, any input results in a unique output.
So, the implication made in the steps shown, albeit useful here, has nothing to do with the fact that they are logarithms of some values, but rather that for any $a,b$,
$$a=b \Rightarrow 6^a=6^b$$
Shubhav Jain
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a = b => f(a) = f(b) seems more like a statement that a function produces a deterministic output. Although unique output is logically correct. – Fomalhaut Nov 03 '23 at 06:42
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If f(a) is not equal to f(b) when a = b, then for the same value input to the function, different(not unique) outputs would be obtained. So, I do think the statement means that a function produces a unique output. However, I would like to understand how deterministic output differentiates from unique output. – Shubhav Jain Nov 03 '23 at 15:42
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If function means what the english word function means (that is, a process) then the only way for a function to produce two different outputs given the same input is if it is non-deterministic and sensitive to context. – Fomalhaut Nov 04 '23 at 00:33
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No, a function has a very specific meaning in Mathematics. Functions are subsets of relations. A relation can have a single element of domain 'related' to multiple elements of the co-domain, and even that is deterministic. When someone writes a=b⇒f(a)=f(b) for a function, this only differentiates functions from such relations (called the one-many relations). Determinism does not seem to have any say in the matter, because there is nothing non-deterministic. – Shubhav Jain Nov 04 '23 at 06:50
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Then it's a very bad name. Because function means doing. When we say "what's its function?" that is logically equivalent to "what does it do?" Names always mean something. – Fomalhaut Nov 04 '23 at 07:30
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See, it doesn't matter. A function is popularly defined in Mathematics. Once we define something, we go along with it. The only thing that is up for debate is if you commented here based on your knowledge of language or mathematics. – Shubhav Jain Nov 04 '23 at 16:21
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Besides, if you wish to see a mathematical function as a process, you can. For example, here, the exponential function, (reducing the definition to the domain of natural numbers), is the process of repeated multiplication a certain number of times. – Shubhav Jain Nov 04 '23 at 16:21
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The issue you have brought up here is with non-determinism.
A well-defined function (or a process) will have a deterministic output.
– Shubhav Jain Nov 04 '23 at 16:23 -
The point I am making is that in mathematics, there are concepts like relations, where it is not necessary for it to be non-deterministic to negate the implication a=b⇒f(a)=f(b). The non-uniqueness of the image in a relation may negate the implication, and that is the only possibility for such a thing to happen. – Shubhav Jain Nov 04 '23 at 16:39
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It is perhaps easier to visualise in terms of natural logarithms. For example:
$\ln(2x-3) = \ln(4) \rightarrow \exp(\ln(2x-3)) = exp(\ln 4) $
$\rightarrow e^{\ln(2x-3)} = e^{\ln(4)} $
Just for fun and interest, we could use a little known "rule" that
$a^{\log b} = b^{\log a}$ where $a,b \neq 0,1$.
Start with $2x-3 = 4$
$\rightarrow (2x-3)^{log_6 7} = 4^{log_6 7}$
This is still valid because I have only raised both sides to the same power.
Using the "rule" $\rightarrow 7^{log_6 (2x-3)} = 7^{log_6 (4)}$
Notice that the radix is 7 and not 6 but it still works.
Nothing special about 7 either. It could be any positive real number besides 0 or 1. The whole mess can undone by applying the "rule" again. All unnecessarily complicated of course but it might be of interest ;-)
The most sensible method is simply:
$\log_6(2x-3) = \log_6(4) \rightarrow (2x-3) = (4)$
For completeness, if $a = b$ then $x^a = x^b$ is perfectly valid and as someone else mentioned, has nothing to do with the properties of logarithms.
KDP
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