Asked by jacĸѕon
ѕolve
хy=80
logх-2logy=1
хy=80
logх-2logy=1
Answers
Answered by
bobpursley
taking logs of first equation:
logx + logy=log80
now subtracting that from the second initial equation
-3logy=1-log80
note that 1=log10
1/y^3=10/80
y= cubrt(8)=2
check my work
logx + logy=log80
now subtracting that from the second initial equation
-3logy=1-log80
note that 1=log10
1/y^3=10/80
y= cubrt(8)=2
check my work
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