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...
