solve log9(x-y)=log3(x+y

Answer 1

logb(M) = logc(M) logb(c)
so
log3(M) = log9(M) log3(9)
log3(x-y) = log9(x-y) log3(9)
but 3^log3(9) = 9 so log3(9)= 2
so
log9(x-y) = (1/2) log3 (x-y)
then
(1/2) log3(x-y) = log3(x+y)
log3(x-y) = log3(x+y)^2
so
x-y = x^2 +2xy+y^2 etc

Answer 2

let log9(x-y) = a ----> 9^a = x-y
then log3(x+y) = a ---> 3^a = x+y

from 9^a = x-y
(3^2)^a = x-y
(3^a)^2 = x-y
3^a = √(x-y)

Thus:
x+y = √(x-y)
(x+y)^2 = x-y

The solutions that appear obvious are:
x = 0, y = -1
x = 1, y = 0

But, I squared the equation, so all answers must be verified:

check: if x=0, y = -1
LS = log9(0) which is not defined
RS = log3(1) = 0
LS ≠ RS

if x = 1, y = 0
LS = log9(1) = 0
RS = log3(1) = 0
LS = RS

so x = 1, y = 0

Answer 3

Here is the graph of (x+y)^2 = x-y
http://www.wolframalpha.com/input/?i=plot+(x%2By)%5E2+%3D+x-y