Simple logarithm question

lostlad

I just forgot how to do them:o

Thanks

Log(x-1) + log(x+1) = 2log(x+2) Ans = [-1r1r4]

hivizman

This question is by no means simple.

Faced with an equation involving logs, the natural thing to do is to express the equation without logs. Adding the logs of two numbers then becomes multiplying the two numbers and multiplying the log of a number by another number becomes raising the first number to the power of the second number.

So log(x-1) + log(x+1) = 2log(x+2) becomes

(x-1)(x+1) = (x+2)^2

Expanding both sides, we get

x^2 - 1 = x^2 +4x +4

And a bit of rearrangement gives

4x = -5, or x = -5/4

But now things get more complex, literally.

Over the real numbers, the log function is defined only for positive values of x. That means that log(-9/4) and log(-1/4), the two terms on the left hand side of the equation, are not actually defined.

Logs of negative numbers exist only when we expand from real to complex numbers. To make matters worse, logs of negative numbers do not have unique values.

To see this, we need to make use of Euler's formula:

exp(ix) = cosx +isinx, where i is the imaginary square root of -1 and exp() is the exponential function. Where x = π (pi), then cosx = -1 and sinx = 0, so exp(iπ) = -1, and thus iπ = log(-1). But cosx = -1 and sinx = 0 for x = -π, and indeed for any odd multiple of π.

Given that -9/4 can be expressed as -1 x 9/4, we can rewrite log(-9/4) as log(-1) + log(9/4), and similarly log(-1/4) as log(-1) + log (1/4), so the equation becomes:

log(-1) + log(9/4) + log(-1) + log(1/4) = 2log(3/4)

Choosing values of log(-1) that cancel out (e.g. +iπ for the first log(-1) term and -iπ for the second log(-1) term), this reduces to

log(9/4) + log(1/4) = 2log(3/4), which is equivalent to

9/4 x 1/4 = 9/16 = (3/4)^2

Numerically, we have:

0.3522 - 0.6020 = -0.2498 = 2 x -0.1249

CJC86

hivizman wrote: »

... so the equation becomes:

log(-1) + log(9/4) + log(-1) + log(1/4) = 2log(3/4)

Choosing values of log(-1) that cancel out (e.g. +iπ for the first log(-1) term and -iπ for the second log(-1) term), this reduces to

log(9/4) + log(1/4) = 2log(3/4), which is equivalent to

9/4 x 1/4 = 9/16 = (3/4)^2

Numerically, we have:

0.3522 - 0.6020 = -0.2498 = 2 x -0.1249

Everything you did up to here is fine, but you can't possibly "choose values of log(-1) that cancel out", since you would then be using two different definitions of log in the same equation! As you said, log is not a single-valued function over the complex numbers, so then you have to choose a consistent definition of log for your means.

Of course you got values 9/4, 1/4 and (3/4)^2 which worked out in your final equation, but there is no x that would give these 3 figures.

The point then, once you get x=-5/4, you check to see if this is a valid answer, as this is the only possibility from our first equation above. It is not a valid answer, because log is not defined for -(1/4) and -(9/4) so there is no solution for x.

hivizman

Perhaps a way to argue it is to note that log(-1) + log(-1) = log(-1 x -1) = log(1) = 0. But all of this assumes that we can manipulate logs of negative numbers in the same way as we can manipulate logs of positive numbers. If we can't, then the false step in my argument is claiming that log(-9/4) = log(-1 x 9/4) = log(-1) + log(9/4). But then, given that x-1 and x+1 turn out to be negative, can we legitimately claim that log(x-1) + log(x+1) = log((x-1)(x+1))? Do we have to assume that log(x-1) and log(x+1) are well-defined to prove that they are not well-defined if the given equation holds? Possibly there's a reductio ad absurdum argument here: for there to be a valid solution to the equation, we must assume that all the log terms are well-defined, which implies that x-1 and x+1 are both positive (a fortiori x+2 will be positive if x+1 is positive). Then the equation implies that x=-5/4, so both x-1 and x+1 are negative, which produces a contradiction. Hence there is no valid solution to the equation.

It would be interesting to know the context of the original question - did whoever set the question realise that getting negative values for x-1 and x+1 raises deeper issues involving the range of values for which the logarithmic function is defined?

CJC86

How's your Complex Analysis? The issue with saying log(-1)+log(-1)=log(1)=0 is that log is only a single valued function on a (chosen) open branch of the complex plane that sweeps out an angle of 2pi. We cannot jump from one side of the branch to the other (this, by the way, is the reason that contour integration works).

If you take log in the general complex analysis sense, then it defines an equivalence class of values. In this sense it is always ok to say that logA+logB=log(AB), but if you are taking log to be a function, then it is only ok if the argument of AB, ie. argA+argB, is in the same branch.

Given the question that was asked, I believe the answer is that no x satisfies the first equation. However, if the question was: "Find a continuous branch of log on which we could find a solution to this equation", then we could give an answer, and the solution would be x=-5/4, but it involves defining log(-1)=i*pi and log(1)=2i*pi.

Long story short OP, the answer to the question is that there is no solution. If your teacher says -5/4 is the solution, point out that log is not defined for negative real numbers.

lostlad

Thanks for the replys the questions was in a book called Technician mathematics volume two and I think it assumed you had no knowledge of complex numbers.

Only had a quick look through what was mentioned there will have a deeper look at it tomorrow..I guess its time to brush up on complex numbers next:)