To solve the log equation 0.5log(2x-1) + log√(x-9) = 1, first rewrite the square root as an exponent and express 1 as log(10), resulting in 0.5log(2x-1) + 0.5log(x-9) = log(10). By multiplying both sides by 2, the equation simplifies to log((2x-1)(x-9)) = log(100). Taking the antilog leads to the equation (2x-1)(x-9) = 100, which expands to 2x^2 - 19x - 91 = 0. The solutions for x are 13 and -3.5.
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