Question

Find the values of x for which logx(1-x^2) is real and defined

Answer 1

x ∈ (0, 1)

Answer 2

For the expression $\log_x(1-x^2)$ to be real and defined, the following conditions must be satisfied:

1. The base of the logarithm must be positive: $x > 0$.
2. The base of the logarithm must not be equal to 1: $x \neq 1$.
3. The argument of the logarithm must be positive: $1-x^2 > 0$.

Let's solve the inequality $1-x^2 > 0$: $1 > x^2$ $x^2 < 1$

This inequality is satisfied when $-1 < x < 1$.

Now we need to find the values of $x$ that satisfy all three conditions:

• $x > 0$
• $x \neq 1$
• $-1 < x < 1$

Combining the conditions $x > 0$ and $-1 < x < 1$, we look for the intersection of the intervals $(0, \infty)$ and $(-1, 1)$. The intersection is the interval $(0, 1)$.

The condition $x \neq 1$ is already satisfied for all values of $x$ in the interval $(0, 1)$, as 1 is not included in this interval.

Therefore, the values of $x$ for which $\log_x(1-x^2)$ is real and defined are the values in the interval $(0, 1)$.