The system to solve:
(1) log10(2000xy)−log10xlog10y=4
(2) log10(2yz)−log10ylog10z=1
(3) log10(zx)−log10zlog10x=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):
log10z+log10x−log10zlog10x=0
log10z(1−log10x)+log10x=0
(4) log10z=−log10x1−log10x
From equation (2):
log102+log10y+log10z−log10ylog10z=1
(5) log102+log10y+log10z(1−log10y)=1
Substitute equation (4) into equation (5):
log102+log10y−log10x1−log10x(1−log10y)=1
(6) log102+log10y(1+log10x1−log10x)−log10x1−log10x=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):
log102000+log10x+log10y−log10xlog10y=4
log102000+log10x+log10y(1−log10x)=4
log10y(1−log10x)=4−log102000−log10x
Note that log102000=log102+log101000=log102+3.
(7) log10y=1−log102−log10x1−log10x
Substitute equation (7) into equation (6) and simplify:
log102+1−log102−log10x1−log10x(1+log10x1−log10x)−log10x1−log10x=1
log102+1−log102−log10x1−log10x11−log10x−log10x1−log10x=1
1−log102−log10x(1−log10x)2−log10x1−log10x=1−log102
We now have a single equation in a single variable, x.
Solve for x
Multiply through by the largest denominator and then simplify:
1−log102−log10x−log10x(1−log10x)=(1−log102)(1−log10x)2
1−log102−2log10x+(log10x)2=(1−log102)(1−2log10x+(log10x)2)
1−log102−2log10x+(log10x)2=1−log102−2(1−log102)log10x+(1−log102)(log10x)2
0=2log102log10x−log102(log10x)2
0=2log10x−(log10x)2
Factorise:
0=log10x(2−log10x)
So we can now solve for x:
log10x=0 or 2. So x=100=1 or 102=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 log10x into equation (7):
Solve for y and z
First root (log10x=0):
log10y=1−log102
log10y=log1010−log102=log10(102)=log105
y=5
Second root (log10x=2):
log10y=1−log102−21−2
log10y=1+log102=log1010+log102=log1020
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 (log10x=0):
log10z=0
z=1
Second root (log10x=2):
log10z=−21−2
log10z=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):
log1010000−log101log105=4
log1010−log105log101=1
log101−log101log101=0
4−0=4
1−0=1
0−0=0
Checks out...
Second solution set (x,y,z)=(100,20,100):
log104000000−log10100log1020=4
log104000−log1020log10100=1
log1010000−log10100log10100=0
6+2log102−2(1+log102)=4
3+2log102−2(1+log102)=1
4−2⋅2=0
6−2=4
3−2=1
4−4=0
Also checks out...
