How to take the log of an equation?

Why did my use of log properties not work out?

• I'm told to find the inverse of [math]y=ln(x+3)[/math]. I flip x and y, and as such, [math]x=ln(y+3)[/math]; here I see two different ways to go about it, the first being [math]ex=e ln(y+3)[/math], --> [math]y=ex-3[/math]; and the second being [math]x^e=ln((y+3)^e)[/math], --> [math]y=(x^e)-3[/math]. From this, it seems that [math]ex=x^e[/math], which clearly isn't right. I'm wondering where I went wrong here. My own messing around with this has brought me to believe the error is likely has to do with an improper use of exponents, and the fact that I'm raising a quantity other than a single variable to an exponent. But if this is the case, I'm curious to know if or how sums of variables, coefficients, etc. can be manipulated [i]appropriately[/i] within the context of the log of an exponent being equal to itself, or vice versa, etc. Edit: it was pointed out that a simpler solution to the equation would be to take the exponential of both sides, [math]e^x=y+3[/math], resulting in [math]y=e^x-3[/math]. This is correct, and much more efficient, but what I'm trying to understand is a seeming discontinuity in log properties per the ways I was attempting to solve it. I could use other examples, but this was just the closest at hand.

• Answer:

  I don't know what you're asking exactly, but many of the properties of things like exponents and logarithms you see in school only work for particular inputs.  For instance, negative numbers do not have logarithms in the real sense.  You can remedy this to some extent working over the complex numbers, but the complex exponential function is not one-to-one so it doesn't have an actual inverse function.  Thus defining a complex logarithm requires making a choice of branch, which breaks any properties that rely on the uniqueness of the inverse. EDIT: Originally there were no question details. Let's invert [math]y = \log(x + 3)[/math] over the real numbers.  Taking the exponential of both sides, [math]e^y = x + 3[/math].  Solving for [math]x[/math], [math]x = e^y - 3[/math].  Done.