The expression log55x2 can be rewritten using logarithmic properties. Initially, it was attempted to express it as 2log55x, but this was incorrect. The correct approach involves separating the logarithm into a sum: log55 + log5x2. This simplifies further to 1 + 2log5x, providing the final rewritten expression. The discussion emphasizes the importance of understanding logarithmic addition and substitution for accurate transformations.
#1
nlsed
10
0
Homework Statement
Rewrite the expression as a sum, difference or multiple of logs.
That's not the right first, step. What do you know about the addition of logs? If that doesn't help, substitute a = 5 and b = x2 and re-ask my question.
The Bob
#3
nlsed
10
0
ooh~
log55 + log5x2 =
1+log5x2 =
1 + 2log5x2!
thank you so much
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