Different ways of expressing logarithms

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The discussion focuses on demonstrating that 2log(e) equals 1/ln(2) using logarithmic laws. Participants suggest using the equation ^{a}log(x) = ^{a}log(b)·^{b}log(x) to manipulate the expression, but one user expresses a desire to solve the problem without relying on this equation. Clarifications about logarithmic notation are provided, emphasizing the correct formatting for base-a logarithms. The conversation highlights the importance of understanding logarithmic properties to derive the desired result. Overall, the thread revolves around finding a proof through manipulation of logarithmic identities.
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Homework Statement


Show that ^{2}log(e)=\frac{1}{ln2}

Homework Equations


^{a}log(x) = ^{a}log(b)\cdot ^{b}log(x)^{a}log(x)=\frac{^{b}log(x)}{^{b}log(a)}

The Attempt at a Solution


How can this be shown? I assume it can be done just using logarithm laws, but I don't see how. I tried manipulating around the factors, but I can't get to it. Can someone please show some way to show what the question asks.
 
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I think you wrongly formatted log_a(x) as a log(x) instead.

You can use the second relevant equation you have noted above. What can you use as the value of 'b' in that equation to get something that could look like the RHS of your desired proof ? And what is log_k(k) for any k in a valid domain?
 
Infinitum said:
I think you wrongly formatted log_a(x) as a log(x) instead.
Where?

Infinitum said:
You can use the second relevant equation you have noted above. What can you use as the value of 'b' in that equation to get something that could look like the RHS of your desired proof ? And what is log_k(k) for any k in a valid domain?
Actually I shouldn't have written that equation. I want to know how to do it, without using this equation. The point of the exercise is actually to make an example which leads to this equation. Is this possible?
 
Maxo said:

Homework Statement


Show that ^{2}log(e)=\frac{1}{ln2}

Homework Equations


^{a}log(x) = ^{a}log(b)\cdot ^{b}log(x)^{a}log(x)=\frac{^{b}log(x)}{^{b}log(a)}

The Attempt at a Solution


How can this be shown? I assume it can be done just using logarithm laws, but I don't see how. I tried manipulating around the factors, but I can't get to it. Can someone please show some way to show what the question asks.

The standard notation for the base-a logarithm of b is ##\log_a b## or ##\log_a(b)##, not ##{}^a \log b## or ##{}^a \log(b)##. Anyway, ##\log_a b = x## means that ##a^x = b##, so you should be able get everything you need from that.
 
You could prove the relevant equation, or the question itself.

Let log_b(x) = p, and log_a(x) = q
Substituting for x in the second equation, by using the definition of logarithm on the first, you should be able to prove the relevant equation.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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