Ln(9/4) + ln (16/9) - ln (3/1)

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The discussion focuses on the correct application of logarithmic properties to simplify the expression ln(9/4) + ln(16/9) - ln(3/1). It highlights that the order of operations matters in logarithmic equations, as incorrect grouping leads to wrong results. The correct simplification results in ln(4/3), while an incorrect approach yields an erroneous ln(144/108). Participants acknowledge common mistakes in handling logarithms and emphasize the importance of recognizing factors early in the simplification process. Understanding these principles is crucial for accurate logarithmic calculations.
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Homework Statement



If you have, for example, 2 + 4 - 1, you can get the answer (5), by doing both:

= 2 + (4 - 1)

and,

= (2 + 4) - 1


But the same logic does not work with logs: to get the right answer (4/3) here you must do:

=(ln(9/4) + ln (16/9)) - ln (3/1)

and NOT:

ln(9/4) + (ln (16/9) - ln (3/1))


Why exactly is this so?



Homework Equations



ln (9/4) + ln (16/9) - ln (3/1)

2. Attempt at a solution

(ln (9/4) + ln (16/9)) - ln (3/1)

= (ln (4) - ln (3))
= ln (4/3) = correct

(ln (9/4) + (ln (16/9)) - ln (3/1))

= ln (9/4) + ln (16/27)
= ln (144/108) = wrong
 
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939 said:
= ln (9/4) + ln (16/27)
= ln (144/108) = wrong

What does that reduce to in lowest terms?
 
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scurty said:
What does that reduce to in lowest terms?

Ahhhh, such a stupid mistake. Thanks.
 
939 said:
Ahhhh, such a stupid mistake. Thanks.

It happens to everybody, you're welcome.

Rather than simplify at the end you can do it earlier by recognizing that 9 divides into 27 evenly and 4 divides into 16 evenly: $$\frac{9}{4} \cdot \frac{16}{27} = \frac{16}{4} \cdot \frac{9}{27}$$.
 
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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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