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How do you solve a system with one equation in the system containing logs?
The two equations in the system are
log x-log y=1 and
x+2y=20
How would you find the answer?
4 Answers
- AlexLv 61 decade agoFavorite Answer
Use: log a - log b = log (a/b)
log x - log y = 1
log (x/y) = 1 = log 10
=> x/y = 10 => x = 10y
x + 2y = 20
10y + 2y = 20
12y = 20
y = 20/12 = 5/3
x = 50/3
- RayLv 71 decade ago
Since log(a)-log(b)=log(a/b) we get log(x)-log(y)=1 or log(x/y)=1 and x/y=10 since log(10)=1.
So x=10y is what we get. Now we also have x+2y=20.
Substitute 10y in for x in equ#2 we get 10y+2y=20 and y=20/12 or 5/3.
Now x+2(5/3)=20 and x=20-10/3 or 50/3.
- 1 decade ago
Plug in "1" for "Y", then multiply 2 and 1 then subtract the product from 20 and you get X=18
