SOLUTION: Solve the equations logx+logy=4 logx+2logy=3 THANK YOU

Algebra ->  Equations -> SOLUTION: Solve the equations logx+logy=4 logx+2logy=3 THANK YOU      Log On


   

Question 995769: Solve the equations
logx+logy=4
logx+2logy=3
THANK YOU
Found 2 solutions by ikleyn, MathLover1:
Answer by ikleyn(52798) About Me  (Show Source):
You can put this solution on YOUR website!
.
system%28log%28x%29+%2B+log%28y%29+=+4%2C%0D%0Alog%28x%29+%2B+2%2Alog%28y%29+=+3%29.

Distract the first equation from the second.  You will get

log%28y%29 = 3+-+4 = -1.

Hence,  y = 1%2F10     (if 10 is the base of logarithms).

From this point,  complete the solution yourself.


Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
I assume, the base is ten
log%28x%29%2Blog%28y%29=4..........eq.1
log%28x%29%2B2log%28y%29=3........eq.2
-------------------------subtract 1 from 2
log%28x%29%2B2log%28y%29-%28log%28x%29%2Blog%28y%29%29=3-4
log%28x%29%2B2log%28y%29-log%28x%29-log%28y%29=-1
log%28y%29=-1
y=10%5E-1
highlight%28y=1%2F10%29=> means log%281%2F10%29=-1

log%28x%29%2Blog%281%2F10%29=4..........eq.1
log%28x%29=4-log%281%2F10%29
log%28x%29=4-%28-1%29
log%28x%29=5
x=10%5E5
highlight%28x+=+100000%29