Asked by muleya
Solve simultaneous equation.
log2(x-14y)=3
logx-log(y+1)=1
log2(x-14y)=3
logx-log(y+1)=1
Answers
Answered by
bobpursley
The first equation is easy, if log base 2 is 3, then
x-14y=8
The second equation is the same as
x/(y+1)= 10 or x =10y+10
then
y+10 -14y=8
-13y=-2
and you have y
then x=10(2/13)+10=11.53
y=.154
x-14y=8
The second equation is the same as
x/(y+1)= 10 or x =10y+10
then
y+10 -14y=8
-13y=-2
and you have y
then x=10(2/13)+10=11.53
y=.154
Answered by
Reiny
I noticed a typo in the above solution.
y+10 -14y=8 should be
10y + 10 - 14y = 8
The solution comes out a bit "nicer".
y+10 -14y=8 should be
10y + 10 - 14y = 8
The solution comes out a bit "nicer".
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