Sum-log

The sum of two numbers is 32, the sum of their logarithms (base 10) is 2.2. Determine these numbers.

Result

x =  25.875
y =  6.125

Solution:

x+y=32 logx+logy=2.2  y=32y  logxy=2.2 xy=102.2  x(32x)=102.2 32xx2102.2=0 x232x+158.4893=0  x1,2=b±D2a=32±390.042 x1,2=16±9.87474965525 x1=25.8747496553 x2=6.12525034475x+y= 32 \ \\ \log x + \log y = 2.2 \ \\ \ \\ y = 32-y \ \\ \ \\ \log xy = 2.2 \ \\ xy = 10^{ 2.2 } \ \\ \ \\ x(32-x) = 10^{ 2.2 } \ \\ 32 x-x^2-10^2.2 = 0 \ \\ x^2 -32 x + 158.4893 = 0 \ \\ \ \\ x_{1,2} = \dfrac{ -b \pm \sqrt{ D } }{ 2a } = \dfrac{ 32 \pm \sqrt{ 390.04 } }{ 2 } \ \\ x_{1,2} = 16 \pm 9.87474965525 \ \\ x_{1} = 25.8747496553 \ \\ x_{2} = 6.12525034475
y=((32)(19.7494993105))/(2 (1))=6.125y=(-(-32) - (19.7494993105))/ (2 \cdot \ (1))=6.125



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