Find the solution to the system of equations Log (2000xy) minus log(x).log y=4 Log(2yz)minus log y.logz=1 Log(zx) minus log(z).logx=0 Base of log is 10 everywhere?

1 Answer
Jun 27, 2018

Eliminate variables one by one, then substitute back. We obtain two solution sets for (x,y,z): (1,5,1) and (100,20,100).

Explanation:

The system to solve:
(1) log_(10)(2000xy)-log_(10)xlog_(10)y=4
(2) log_(10)(2yz)-log_(10)ylog_(10)z=1
(3) log_(10)(zx)-log_(10)zlog_(10)x=0

Recall that log(ab)=loga+logb.

Eliminate z

First, eliminate the variable z from our system of three equations in three unknowns.

From equation (3):
log_(10)z+log_(10)x-log_(10)zlog_(10)x=0
log_(10)z(1-log_(10)x)+log_(10)x=0
(4) log_(10)z=-log_(10)x/(1-log_(10)x)

From equation (2):
log_(10)2+log_(10)y+log_(10)z-log_(10)ylog_(10)z=1
(5) log_(10)2+log_(10)y+log_(10)z(1-log_(10)y)=1

Substitute equation (4) into equation (5):

log_(10)2+log_(10)y-log_(10)x/(1-log_(10)x)(1-log_(10)y)=1
(6) log_(10)2+log_(10)y(1+log_(10)x/(1-log_(10)x))-log_(10)x/(1-log_(10)x)=1

We have now eliminated the variable z and have a system of two equations - (1) and (6) - in two unknowns, x and y.

Eliminate y

We now repeat the process to eliminate y.

From equation (1):
log_(10)2000+log_(10)x+log_(10)y-log_(10)xlog_(10)y=4
log_(10)2000+log_(10)x+log_(10)y(1-log_(10)x)=4
log_(10)y(1-log_(10)x)=4-log_(10)2000-log_(10)x
Note that log_(10)2000=log_(10)2+log_(10)1000=log_(10)2+3.
(7) log_(10)y=(1-log_(10)2-log_(10)x)/(1-log_(10)x)

Substitute equation (7) into equation (6) and simplify:
log_(10)2+(1-log_(10)2-log_(10)x)/(1-log_(10)x)(1+log_(10)x/(1-log_(10)x))-log_(10)x/(1-log_(10)x)=1
log_(10)2+(1-log_(10)2-log_(10)x)/(1-log_(10)x) 1/(1-log_(10)x)-log_(10)x/(1-log_(10)x)=1
(1-log_(10)2-log_(10)x)/(1-log_(10)x)^2 -log_(10)x/(1-log_(10)x)=1-log_(10)2

We now have a single equation in a single variable, x.

Solve for x

Multiply through by the largest denominator and then simplify:
1-log_(10)2-log_(10)x -log_(10)x(1-log_(10)x)=(1-log_(10)2)(1-log_(10)x)^2
1-log_(10)2-2log_(10)x+(log_(10)x)^2=(1-log_(10)2)(1-2log_(10)x+(log_(10)x)^2)
1-log_(10)2-2log_(10)x+(log_(10)x)^2=1-log_(10)2-2(1-log_(10)2)log_(10)x+(1-log_(10)2)(log_(10)x)^2
0=2log_(10)2log_(10)x-log_(10)2(log_(10)x)^2
0=2log_(10)x-(log_(10)x)^2
Factorise:
0=log_(10)x(2-log_(10)x)

So we can now solve for x:
log_(10)x=0 or 2. So x=10^0=1 or 10^2=100. As we ended up with a quadratic for x, we have two possible answers. To obtain y and z, we now substitute back through our equation process. Substitute our values for log_(10)x into equation (7):

Solve for y and z

First root (log_(10)x=0):
log_(10)y=1-log_(10)2
log_(10)y=log_(10)10-log_(10)2=log_(10)(10/2)=log_(10)5
y=5

Second root (log_(10)x=2):
log_(10)y=(1-log_(10)2-2)/(1-2)
log_(10)y=1+log_(10)2=log_(10)10+log_(10)2=log_(10)20
y=20

So we now have two solution pairs (x,y): (1,5) and (100,20). We now substitute back into equation (4) to solve for z:

First root (log_(10)x=0):
log_(10)z=0
z=1

Second root (log_(10)x=2):
log_(10)z=-2/(1-2)
log_(10)z=2
z=100

So we end up with two solution sets (x,y,z): (1,5,1) and (100,20,100).

Check our solutions

Double check our solution by substituting both solution sets back into equations (1-3):

First solution set (x,y,z)=(1,5,1):

log_(10)10000-log_(10)1log_(10)5=4
log_(10)10-log_(10)5log_(10)1=1
log_(10)1-log_(10)1log_(10)1=0

4-0=4
1-0=1
0-0=0

Checks out...

Second solution set (x,y,z)=(100,20,100):

log_(10)4000000-log_(10)100log_(10)20=4
log_(10)4000-log_(10)20log_(10)100=1
log_(10)10000-log_(10)100log_(10)100=0

6+2log_(10)2-2(1+log_(10)2)=4
3+2log_(10)2-2(1+log_(10)2)=1
4-2*2=0

6-2=4
3-2=1
4-4=0

Also checks out...