Can u help, cos i cant solve this sum and im getting obsessed?

Question

Solve simultaneously:
log(x-y-1)=0, log(xy)+1=0

Answer 1

10^log(x-y-1) = x-y-1 = 1

10^[(log(xy) + 1] = 10xy = 1

x-y-1 = 1
10xy = 1

y =x-2 into 10xy =1 and get

10x(x-2)=1
10x^2-20x-1 = 0


x1,2 = (20 +- sqrt(400 + 40))/20 = 1 +- sqrt(110)/10

x1 = 1 + sqrt(110)/10
y1 = sqrt(110)/10 - 1

x2 = 1 - sqrt(110)/10
y2 = -1 - sqrt(110)/10

Both are smaller than 0, so their product is positive which is in the log function's domain.

Answer 2

log (x) = 0 is the same as x = 1. So, change the first one x - y - 1 = 1, or x - y = 2.

log(xy) = -1 is the same as xy = 1/10.

You are left with the (much simpler) equations:

x - y = 2
10xy = 1

Use the first to get y = x - 2. Plug this into the second to get:

10x(x-2) = 1

So, it looks like two solutions to me, the two solutions for

10x^2 - 20x - 1 = 0

Also, as Amit (next poster) points out, in one solution the x,y pair are both negative, yielding a positive product for the logarithm. Both solutions are therefore valid. You're right, this one is tricky!

Answer 3

log(x-y-1)=0 so x-y-1=10^0=1 and x-y=2 or x=2+y
log(xy)+1=0 so 10xy = 1
Thus 10(y+2)y = 1
y² +2y = 1/10
y = 0.0488 or -2.0488
so x = 2.0488 or 0.0488

Answer 4

x=0, y=0

Answer 5

I get x = 2+y = 2.0488088482. That makes y=the decimal part.

(I'm assuming base 10 logs.) You can ck:

log(2.0488088482-.0488088482-1) = ? and
log(2.0488088482 x .0488088482) + 1 = ?

(Answers wouldn't let me close my parentheses after the 1 in the next to last line.) BTW you have to discard any solutions that make the log argument negative, since logs of negatives are undefined.

Answer 6

first equation-
x-y-1=e^0=1
x-y=2--------------(a)
2nd equation-
log(xy)=-1
xy=e^(-1)
xy=1/e--------------(b)
(x+y)^2=(x-y)^2+4xy
put values from equqtion (a)and(b)
(x+y)^2=4+4*(1/e)
now you will get value of (x+y)
then solve it with equation(a)
you will get answer.
Good Luck.
you can ask any type of maths problem personally,i'll feel glad to solve that because math is my passion.