solve log9(x-y)=log3(x+y
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
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
Here is the graph of (x+y)^2 = x-y
http://www.wolframalpha.com/input/?i=plot+(x%2By)%5E2+%3D+x-y
To solve the equation log9(x-y) = log3(x+y), we can use the properties of logarithms. Specifically, we can use the property that log base a of b is equal to log base c of b divided by log base c of a.
First, let's convert the equation to its corresponding exponential form. For log9(x-y), the base is 9 and the result is (x-y). Similarly, for log3(x+y), the base is 3 and the result is (x+y).
So, we have:
9^(log9(x-y)) = 9^ (log3(x+y))
By applying the conversion to exponential form, we obtain:
(x-y) = (x+y)^log9(3)
Since we cannot simplify this any further, we can assume that (x+y)^log9(3) is a constant value, denoted as "c".
Therefore, (x-y) = c.
Finally, we can solve for x by isolating it:
x = (c + y)
This is the general solution to the equation log9(x-y) = log3(x+y). The specific value of c will depend on the context of the problem or any additional information provided.