- #1
- 34
- 0
Homework Statement
2log(x-3)^2=log(x-3)+3
Homework Equations
x>3
The Attempt at a Solution
2log(x^2-6x+9)-log(x-3)=3
log(x-3)=3
Not sure where to go from here seeing that it factores and becomes x=3 which makes the solution undefined
You seem to have lost the coefficient of the first log term.Homework Statement
2log(x-3)^2=log(x-3)+3
Homework Equations
x>3
The Attempt at a Solution
2log(x^2-6x+9)-log(x-3)=3
log(x-3)=3
Not sure where to go from here seeing that it factores and becomes x=3 which makes the solution undefined
Using "laws of logarithms", 2log(x-3)^2= 4 log(x- 3) and thenHomework Statement
2log(x-3)^2=log(x-3)+3
Homework Equations
x>3
The Attempt at a Solution
2log(x^2-6x+9)-log(x-3)=3
log(x-3)=3
Not sure where to go from here seeing that it factores and becomes x=3 which makes the solution undefined
That's what I got but it doesn't work. But, treating it as a quadratic, I got [itex]x=10\sqrt{10}+3[/itex], which does work, as well as 13 again. Why is this?If the logarthm is base 10 that says x- 3= 10. If it is base e, x- 3= e.
since a base is missing it is assumed to be base 10Does "log" mean log10 or loge (same as ln). I assumed you meant log10
it is 2log((x-3))^2 i just transformed it in the questionIs the expression on the left side 2 log( (x - 3)^2) or 2 (log(x-3))^2? IOW, what is being squared
That is exactly what you wrote originally. It does NOT answer Mark44's question and does not make sense: "[itex]f(x)^2[/itex]" is meaningless unless you interpret it as [itex](f(x))^2[/itex]. But in your original question, you wrote as log(x^2- 6x+ 9) which is [itex]ln((x-3)^2)[/itex] and could be more easily written as [itex]2ln(x-3)[/itex].since a base is missing it is assumed to be base 10
it is 2log((x-3))^2 i just transformed it in the question