Asked by value
Logx + logy=4 logx +2logy=3
Answers
Answered by
Reiny
from what you have:
Logx + logy=4 logx +2logy=3
log(xy) = log(x^4 y^2) = 3
so : log(xy) = log(x^4 y^2)
x^4 y^2 = xy
x^3 y = 1
y = 1/x^3 , of course x > 0
in logx + logy = 3
logx + log(1/x^3) = 3
log(x(1/x^3)) = 3
log (1/x^2) = 3
1/x^2 = 10^3 = 1000
1000x^2 = 1
x^2 = 1/1000
x = 1/10√10 = √10/100 or appr .03162
y = 1/x^3 = 10000√10 or appr 31622.8
Logx + logy=4 logx +2logy=3
log(xy) = log(x^4 y^2) = 3
so : log(xy) = log(x^4 y^2)
x^4 y^2 = xy
x^3 y = 1
y = 1/x^3 , of course x > 0
in logx + logy = 3
logx + log(1/x^3) = 3
log(x(1/x^3)) = 3
log (1/x^2) = 3
1/x^2 = 10^3 = 1000
1000x^2 = 1
x^2 = 1/1000
x = 1/10√10 = √10/100 or appr .03162
y = 1/x^3 = 10000√10 or appr 31622.8
Answered by
bobpursley
logx+logy=4logx+2logy=3 ??? is the first equals sign a + ? If so, then
log(x*y*x^4*y^2=3
log (x^5 y^3)=3 and you can take the antilog
x^5y^3=1000
log(x*y*x^4*y^2=3
log (x^5 y^3)=3 and you can take the antilog
x^5y^3=1000
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