What if X is 0 or lower in Logarithm Cancellation Law

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The discussion centers on the implications of the logarithm cancellation law when x is 0 or lower. It is emphasized that log(x) is undefined for values of x less than or equal to 0, rendering any logarithmic laws inapplicable in such cases. Participants highlight that this fundamental property of logarithms leads to a lack of meaningful results when x is non-positive. The conversation reiterates that any mathematical operations involving logarithms must adhere to this restriction. Thus, the topic underscores the importance of understanding the domain of logarithmic functions in mathematical contexts.
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What difference will it make?
 
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SELFMADE said:
What difference will it make?

Do you have an example of what you are talking about?
 
log(x) is undefined for x<=0, so any laws involving logs do not make sense when x<=0.
 
Dunkle said:
log(x) is undefined for x<=0, so any laws involving logs do not make sense when x<=0.

That does it. Thanks
 
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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