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# log equation!!!!!!

eh guys, who can solve this?

solve the simultaneous equation:

$$\log{x} +\log{y} = 1$$

$$x+y= 1$$

Note by Samuel Ayinde
2 years, 4 months ago

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logx + logy can be written as log(xy). which can be further written as log(x(1-x)), using x+y=1 .Suppose the base of the log is 'a' then x(1-x)=a which is quadratic in nature and hence can be solved where x=(1+sqrt(1-4a))/2,(1-sqrt(1-4a))/2 and then pluging x in x+y=1 gives y =(1-sqrt(1-4a))/2,(1+sqrt(1-4a))/2,(1-sqrt(1-4a))/2 respectively! Further to get real solutions 0<a<=1/4. · 2 years, 4 months ago