How do you find the interval of existence for the real function $ln (1 + \frac{x}{ln (1 - x)})$ $ln(1+x/(ln(1-x)))$ ?

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How do you find the interval of existence for the real function $ln (1 + \frac{x}{ln (1 - x)})$ $ln(1+x/(ln(1-x)))$ ?

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Answer 1 · 2016-06-21T22:09:28

The domain is $(0, 1)$ $(0, 1)$

Explanation:

The domain of the Real valued function of Reals $ln (x)$ $ln(x)$ is $(0, \infty)$ $(0, oo)$

So looking at the denominator $ln (1 - x)$ $ln(1-x)$ first, we require:

$1 - x > 0$ $1 - x > 0$ .

Add $x$ $x$ to both sides to find $1 > x$ $1 > x$ , that is $x < 1$ $x < 1$

Additionally we require $ln (1 - x) \neq 0$ $ln(1-x) != 0$ in order that the denominator be non-zero. Hence $x \neq 0$ $x != 0$ .

So $x \in (- \infty, 0) \cup (0, 1)$ $x in (-oo, 0) uu (0, 1)$

In order that $1 + \frac{x}{ln (1 - x)} > 0$ $1+x/(ln(1-x)) > 0$ , we require:

$\frac{x}{ln (1 - x)} > - 1$ $x/(ln(1-x)) > -1$

Split into cases:

Case $x \in (0, 1)$ $x in (0, 1)$

Then $1 - x \in (0, 1)$ $1-x in (0, 1)$ and $ln (1 - x) \in (- \infty, 0)$ $ln(1-x) in (-oo, 0)$

Multiply both sides of the inequality by $ln (1 - x)$ $ln(1-x)$ and reverse the inequality (since $ln (1 - x) < 0$ $ln(1-x) < 0$ ) to get:

$x < - ln (1 - x)$ $x < -ln(1-x)$

This inequality holds in the whole interval $(0, 1)$ $(0, 1)$ . I may prove later, but here's a plot for now:

graph{(y+ln(1-x))(y-x)=0 [-1.895, 3.105, -0.68, 1.818]}

Case $x \in (- \infty, 0)$ $x in (-oo, 0)$

In this case:

$1 - x > 1$ $1 - x > 1$

So:

$ln (1 - x) > 0$ $ln(1-x) > 0$

Multiply both sides of the required inequality by $ln (1 - x)$ $ln(1-x)$ to get:

$x > - ln (1 - x)$ $x > -ln(1-x)$

This is false for all $x < 0$ $x < 0$ so there is no $x < 0$ $x < 0$ in the domain.

Proof

What is the slope of $- ln (1 - x)$ $-ln(1-x)$ ?

$\frac{d}{d x} - ln (1 - x) = \frac{1}{1 - x}$ $d/(dx) -ln(1-x) = 1/(1-x)$

So when $x \in (0, 1)$ $x in (0, 1)$ we find:

$\frac{1}{1 - x} > 1$ $1/(1-x) > 1$

and when $x \in (- \infty, 0)$ $x in (-oo, 0)$ we find:

$\frac{1}{1 - x} < 1$ $1/(1-x) < 1$

When $x = 0$ $x = 0$ , $- ln (1 - x) = - ln (1) = 0$ $-ln(1-x) = -ln(1) = 0$

So $- ln (1 - x)$ $-ln(1-x)$ is steeper than $y = x$ $y=x$ when $x > 0$ $x > 0$ and less steep when $x < 0$ $x < 0$ . They both pass through $(0, 0)$ $(0, 0)$ .

Answer 2 · 2016-06-21T22:53:32

${log}_{e} (1 + \frac{x}{{log}_{e} (1 - x)})$ $log_e(1+x/(log_e(1-x)))$

is feasible only for $0 < x < 1$ $0 < x < 1$

Explanation:

${log}_{e} (1 + \frac{x}{{log}_{e} (1 - x)}) = {log}_{e} (\frac{{log}_{e} (1 - x) + x}{{log}_{e} (1 - x)})$ $log_e(1+x/(log_e(1-x)))=log_e((log_e(1-x)+x)/(log_e(1-x)))$

The feasibility condition is

$\frac{{log}_{e} (1 - x) + x}{{log}_{e} (1 - x)} > 0$ $(log_e(1-x)+x)/(log_e(1-x)) > 0$

$1)$ $1)$ For $0 < x < 1$ $0 < x < 1$ we have the condition

${log}_{e} (1 - x) + x < 0 \to {log}_{e} (\frac{1}{e^{x} (1 - x)}) > 0$ $log_e(1-x)+x < 0->log_e (1/(e^x(1-x))) > 0$

or $\frac{1}{e^{x} (1 - x)} > 1 \to e^{- x} > 1 - x$ $1/(e^x(1-x)) > 1->e^{-x} > 1-x$

but $y = 1 - x$ $y=1-x$ is tangent to $y = e^{- x}$ $y=e^{-x}$ at $x = 0$ $x = 0$ and truly

$1-x <= e^{-x} forall x in RR$

$2)$ Considering now the condition $1 < x < oo$ , the feasibility condition is

$log_e(1-x)+x > 0->log_e(e^x(1-x)) > 0$

or

$1-x>e^{-x}$

but $y=1-x$ is tangent to $y=e^{-x}$ at $x = 0$ and truly

$1-x <= e^{-x} forall x in RR$

$3)$ For $-oo > x > 0$ we have

$1+x/log_e(1+x) < 0$ because $delta/log_e(1+delta) >=1$

Concluding,

$log_e(1+x/(log_e(1-x)))$

is feasible only for $0 < x < 1$

Answer 3 · 2016-06-22T04:26:47

$( 0, 1 )$ .

Explanation:

$ln(1-x)$ exists, for x in [-1, 1)#.

$ln(1+x/(ln(1-x)))$ exists, for $x/(ln(1-x)) in (-1, 1]$ .

Now, as $x to 0$ , the given function $to -oo$ .

The proof is a tribute to L'Hospital rule..

For $x < 0, ln(1-x)>0 and x/ln(1-x)<-1$ .

So, the argument for the given ln function becomes negative. The

function does not exist.

Thus, the answer is (0, 1).. .

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