Question

When finding the domain of log5(1+x/1-x), you get the inequality -1 12 years ago

Answer 1

You can re-write it this way
\[y = \log_5(1 + x) - \log_5(1 - x)\]

Two things:

We know that \(x \ne 1\) because if that were the case, that would make the denominator zero and the fraction undefined.

Secondly, if \(x = -1\), then we would have \(\log_5{0}\) which is not possible because let's suppose \(\log_5{0} = y\):

Then \(5^y = 0\) But there is no such y that would make this true. When using exponents, whether the exponent is positive or negative, you can only gain a positive result from it.

And obviously if \(x < -1\), then you would end up with the log of a negative number. We cannot take the log of a negative number and get a real number result.

If \(x > 1\) again we end up with the log of a negative number.

This restricts the domain to \(-1 < x < 1\)

Answer 2

@ineffectivejt

Answer 3

awesome man thank you!