How do I solve for x in this log equation?

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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.
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Homework Statement


0.5log(2x-1)+log\sqrt{x-9}=1


Homework Equations


N/A


The Attempt at a Solution


I have no idea how to approach this.
 
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use this laws:
alogb=logb^{a}
logm+logn=logmn
 
first, rewrite the square root as an exponent, and rewrite 1 as log(10), so you get

0.5log(2x-1) + log(x-9)^(1/2) = log(10)

Bring the exponent down to get 0.5log(x-9)

0.5log(2x-1) + 0.5log(x-9) = log(10)

multiply both sides by 2

log(2x-1) + log(x-9) = 2log(10)

Raise the 10 in the log function to the 2 (100) and multiply the two inner log functions

log[(2x-1)(x-9)] = log(100)

take the antilog of both sides and solve for x

(2x-1)(x-9) = 100

2x^2 - 19x - 91

X=13
X= -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
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