In terms of Briggs logarithms, the solution for x2 can be written as x2 = -log(6)/log(3) ≈ -1.63093 (A = log(6) = 0.77815, B = log(3 )= 0.47712) (Standard Briggs tables to five decimal places.) In fact, the engineers did not divide these two numbers, but re-logarithmized the quotient and then anti-logarithmized the resulting difference. A/B=10^(log(A)-log(B)). log(A/B)=log(10A)-log(10B) log(10A)=0.89106, log(10B)=0.67863. log(log(6)) - log(log(3)) = 0.21244 10^0.21244 = 1.63093. That's how it was calculated in the pre-computer era.😎
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In terms of Briggs logarithms, the solution for x2 can be written as x2 = -log(6)/log(3) ≈ -1.63093 (A = log(6) = 0.77815, B = log(3 )= 0.47712) (Standard Briggs tables to five decimal places.) In fact, the engineers did not divide these two numbers, but re-logarithmized the quotient and then anti-logarithmized the resulting difference. A/B=10^(log(A)-log(B)). log(A/B)=log(10A)-log(10B)
log(10A)=0.89106, log(10B)=0.67863. log(log(6)) - log(log(3)) = 0.21244 10^0.21244 = 1.63093. That's how it was calculated in the pre-computer era.😎
(1) lg[2^x*3^x^2]=lg6 ; (2) x^2*lg(3)+x*lg2-[lg(3)+lg(2) ]=0 . x1=1 , x2=lg(6)/lg(3) { Vietta} .
With respect , Lidiy